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```									The Indeterminate Limit Rule

Jason Schanker (jas42@duke.edu)

15 Lesley Drive

Syosset, NY 11791

Duke University Class of 2005
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About the Author

In his senior year in high school, Jason Schanker became very interested in
mathematical theory and began to examine a number of concepts in calculus. Although
calculus provided him with a broad new repertoire of problems to solve and tools to solve
them with, Jason spent a lot of his time on division by zero and limits involving infinity.
After much time and dedication, he put forward a new theorem to encompass his ideas
regarding limits containing infinity. Following the completion of the report in his senior
year of high school, Jason received assistance from Dr. Kraines at Duke University in the
beginning of his freshman year of college in making revisions to his report and in
submitting his work.
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Abstract

Not all limits in certain indeterminate forms can be solved easily by using
L’Hôpital’s Rule. Specifically, the rule would be difficult to implement on limits of
fractions where the numerator and denominator both approach infinity and both contain
exponential functions. In these cases, differentiating the numerator and denominator to
simplify the limit would generally be futile since the derivatives of these functions
contain the original exponential functions themselves.
To remedy this problem, I have derived and proved a limit rule, which eliminates
these troublesome functions and simplifies the limits into something that can be solved
using L’Hôpital’s Rule. In my report, I state this Indeterminate Limit Rule and prove it.
I then proceed to mention and prove corollaries that make this rule more versatile. In
each of these sections, I conclude with an example using the theorem or corollary that
was most recently stated and proved. I conclude with a section that analyzes the Rule,
investigates the Rule’s inner workings, and examines an important implication of the
theorem.
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Introduction


Some limits with the indeterminate form,           , cannot be easily solved using

L’Hôpital’s Rule. Specifically, the theorem will not be very helpful in solving limits of
this form when they contain exponential functions in both the numerator and
denominator. For example, one application of L’Hôpital’s Rule on the limit,
2.8 x
lim                       would result in a more complex limit of
x  10 x 50 e x ln(10 x)

2.8 x ln 2.8
lim                                                             . Subsequent applications of the Rule
x  10 x 50 e x ln(10 x)  10e x (50 x 49 ln(10 x)  10 x 49 )

would make this limit increasingly more arduous to solve.
The problem with exponential functions is that they are often contained within
their derivatives. Therefore, if these functions are unbounded at the value in question,
their derivatives are also likely to be unbounded at this value. The result of this

consequence is that limits in the indeterminate form ,          , containing these types of

functions in both the numerator and denominator, will most likely stay in this form
regardless of the number of times L’Hôpital’s Rule is applied.
To remedy this problem, I have derived and proved my own theorem that I have
denoted the Indeterminate Limit Rule. This Rule can be used in conjunction with
L’Hôpital’s Rule to solve limits in which the numerator and denominator both have
limiting values of infinity and both contain exponential functions. The Rule
demonstrates that these types of limits can be determined by finding the limiting value of
the ratio of the natural logarithms of the numerator to the denominator as the variable
approaches the same value as it does in the limit in question. If this ratio is less than 1,
then the limiting value is zero. If the ratio’s greater than 1 or infinity, then the limiting
value is infinity. If the ratio is 1, then the test is inconclusive. However, the Limit Rule
should not be necessary in these cases.
The Indeterminate Limit Rule is an indirect extension of L’Hôpital’s Rule.
L’Hôpital’s Rule states that the limiting value of a ratio of two functions with infinite
limits will be equivalent to the limiting value of the ratio of the functions’ derivatives at
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the place where the functions are unbounded. The Indeterminate Limit Rule indirectly
says that the limiting value of a ratio of two functions with infinite limits will be reflected
by the limiting value of the ratio of the ratios of the functions’ derivatives to their
functions. This idea is indirectly stated by the Indeterminate Limit Rule because the Rule
works with the ratio of the natural logarithms of the functions. However, if the limiting
value of a function is positively unbounded, its natural logarithm will also be positively
f ( x)
unbounded. Consequently, if L’Hôpital’s Rule can be applied to lim                                , which is the
x a   g ( x)
                                  ln( f ( x))
limit of the indeterminate form,          , then it can be applied to lim               . Therefore, if
                             x a ln( g ( x ))

f ( x)       ln( f ( x))
the Indeterminate Limit Rule is being applied to lim                         , lim             will be the limit
x a   g ( x) xa ln( g ( x))
f ( x)
that needs to be solved in order to find lim                    , but a single application of L’Hôpital’s
x a   g ( x)
ln( f ( x))                              ln( f ( x))
Rule on lim                  will show that solving lim               is equivalent to solving
x a ln( g ( x ))                        x a ln( g ( x ))

f ' ( x)
(        )
f ( x)
lim              . The reason the Rule is a reflection rather than an equivalency like
x a g ' ( x )
(          )
g ( x)
ln( f ( x))                                       f ( x)
L’Hôpital’s Rule is because lim                     is likely to be closer to 1 than lim          when
x a   ln( g ( x))                                  x a g ( x )

lim f ( x)   and lim g ( x)   . This occurrence is due to the fact that the faster a
xa                  x a

function grows, the greater the growth decrease will be from the function to its natural
logarithm. Consequently, the growth of ln( f ( x)) will be closer to the growth of ln( g ( x))
f ( x)
than the growth of f(x) will be to the growth of g(x). Therefore, lim                            will most
x a   g ( x)
ln( f ( x))
likely be further away from 1 than lim                        and will not be closer.
x a   ln( g ( x))
The Indeterminate Limit Rule works because if an unbounded function, f (x) ,
grows faster than another one unbounded at the same location, g (x) , then the growth of
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f ' ( x)
ln( f ( x)) , which is equivalent to            , will be greater than or equal to the growth of
f ( x)
g ' ( x)
ln( g ( x)) , which is equivalent to            , when x approaches the shared unbounded
g ( x)

location. For example, ln(x) , x n , b x , and x x are all unbounded as x becomes arbitrarily
large. (b and n are constants with b > 1.) However, the rate of growth of all these
functions and hence the values they obtain at any given large x value vary. Specifically,
the above listed functions’ growth rates are in ascending order so ln(x) grows the slowest

and x x grows the fastest as x becomes arbitrarily large. Taking the ratio of each of these
functions’ derivatives to the original functions, which is equivalent to the growth of the
1   n
natural logarithms of these functions, results in           , , ln b , and ln( x)  1 ,
x ln x x
1
respectively.          , which reflects ln(x) , approaches 0 as x becomes arbitrarily large
x ln x
while ln( x)  1 , which reflects x x , becomes arbitrarily large as x becomes arbitrarily
n
large.     will also approach 0 as x becomes arbitrarily large, but it will approach the
x
1
constant slower than           will. ln b is unaffected by x and will have a limiting value
x ln x
n
that falls in between the limiting values of       and ln( x)  1 , exclusively, as x grows
x
without bound.
As illustrated by the above four examples, the Indeterminate Limit Rule preserves
the order of growth rates to the extent that if a function grows faster than another function
at a certain value or at its extremes, then the natural logarithm of the faster-growing
function is guaranteed to not grow slower than the natural logarithm of the slower-
growing function at these locations. For this reason, the limits of the ratio of these
natural logarithms will never be less than 1 if the faster-growing function is in the
numerator and will never exceed 1 if the faster-growing function is in the denominator.
This conclusion comes directly from L’Hôpital’s Rule which ensures that functions with
faster growth and thus greater derivatives will have greater or at least equivalent limiting
values. These premises form the basis of the Indeterminate Limit Rule.
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I. The Indeterminate Limit Rule
Theorem I:

If lim f ( x)   and lim g ( x)   (a can be a finite number or it can represent the end
xa                   x a
behavior of the function, a =   ), then:

ln( f ( x))                  f ( x)
1) If lim                 1 , then lim          0
x  a ln( g ( x ))            x a g ( x)

ln( f ( x))                  f ( x)
2) If lim                 1 , then lim          
x  a ln( g ( x ))            xa g ( x)

ln( f ( x))                    f ( x)
3) If lim                  , then lim           
x  a ln( g ( x ))              xa g ( x)

Proof:

ln( f ( x))
Let L = lim                . If a =  , then for any   0 ,
x a ln( g ( x ))

ln f ( x)
N  0 : x  N , L                     L   and ln g ( x)  0 . If a = –  , then for any   0 ,
ln g ( x)
ln f ( x)
N  0 : x  N , L                     L   and ln g ( x)  0 . Otherwise, if a is finite, there
ln g ( x)
exists a delta such that for all x, where a    x  a   , and  , where   0 ,
ln f ( x)
L               L   and ln g ( x)  0 . These conclusions follow directly from the
ln g ( x)
formal definition of limits. Since lim g ( x)   , an N or a  that satisfy ln g ( x)  0 can
x a

be found by selecting values for which g ( x)  1 along the entire domain. To find an N or
ln f ( x)
a  that simultaneously satisfies L                     L   and ln g ( x)  0 , select the
ln g ( x)
smaller  or the N with the greater absolute value that satisfies both conditions, if a is
finite or infinite, respectively. For all assignments to a, there is some domain for which
ln f ( x)
L                L   and ln g ( x)  0 . Some algebra can be performed on the former
ln g ( x)
inequality:

Multiplying the inequality through by ln g ( x) , the following is obtained (The
inequality signs do not change because ln g ( x) is always positive over the restricted
domain.):

( L   ) ln g ( x)  ln f ( x)  ( L   ) ln g ( x)
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Performing the power logarithm rule on the left and right sides of the inequality

produces the following:

ln( g ( x) L  )  ln f ( x)  ln( g ( x) L  )

Performing the natural logarithm’s inverse on each part of the inequality (Raising e to

each part of the inequality) results in (The inequality signs do not change because the

exponential function is an increasing function over its entire domain.):

g ( x ) L   f ( x )  g ( x ) L  

By dividing the inequality through by g(x), the following results are received:

f ( x)
g ( x) L  1            g ( x) L  1
g ( x)

ln f ( x)
Since the domain for which L                                 L   and ln g ( x)  0 are true is the
ln g ( x)

f ( x)
domain determined by the limit as x approaches a, and g ( x) L  1                          g ( x) L  1
g ( x)

ln f ( x)
is an algebraic manipulation of L                               L   for ln g ( x)  0 , the limit as x
ln g ( x)

approaches a can be applied on the inequality to receive:

f ( x)
lim g ( x) L  1  lim              lim g ( x) L  1 (1)
x a                   x a   g ( x) x a

1) L < 1: When L is less than 1, let  be some number less than 1 – L. This number will

be able to satisfy the condition that  is positive since L is less than 1. With this

assignment to epsilon, L    1 will be negative since L – 1 is negative, and subtracting a

positive number, epsilon, from a negative number will keep the number negative.

L    1 will also be negative since assigning epsilon to 1 – L would make L    1
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zero. However, epsilon has been assigned to a number smaller than 1 – L so L    1

must be smaller than zero. Therefore, the left and right sides both simplify to limits

where g(x) is raised to a negative power. Since lim g ( x)   , raising g(x) to a negative
x a

power in the limit will result in a fraction with a constant numerator and a denominator

that grows without bound. Consequently, the limit will be 0. Using these observations

and the inequality (1), the following is received:

f ( x)
0  lim            0
xa     g ( x)

f ( x)
By the Squeeze Law, we obtain the desired results: lim                    0.
x a   g ( x)

2) L > 1: When L is greater than 1, let  be some number less than L – 1. This number

will be able to satisfy the condition that  is positive since L is greater than 1. With this

assignment to epsilon, L    1 will be positive since L – 1 is positive, and adding a

positive number, epsilon, to another positive number will keep the number positive.

L    1 will also be positive since assigning epsilon to L – 1 would make L    1 zero.

However, epsilon has been assigned to a number smaller than L – 1 so L    1 must be

greater than zero. Therefore, the left and right sides both simplify to limits where g(x) is

raised to a positive power. Since lim g ( x)   , raising g(x) to a positive power in the
x a

limit will result in a function that grows without bound. Consequently, the limit will be

 . Using these observations and the inequality (1), the following is received:

f ( x)
  lim            
x a   g ( x)

f ( x)
By the Squeeze Law, we obtain the desired results: lim                    .
xa    g ( x)
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ln f ( x)
3) L =  : Since by the formal definition of limits,                                                R for any R over the
ln g ( x)

domain determined by the limit, let R = 1. This assignment results in the inequality,

ln f ( x)
 1 . Taking the limit as x approaches a on both sides of the inequality results in:
ln g ( x)

ln( f ( x))                  ln( f ( x))                         f ( x)
lim                 lim 1 or lim                 1 . Since L > 1, lim          . This conclusion
x a   ln( g ( x))   x a     x  a ln( g ( x ))                    xa g ( x)

comes directly from the proof of part 2 of the Indeterminate Limit Rule.

 The three parts of the Indeterminate Limit Rule hold true.

Example I:
1

2 x4
lim                      cos( x 3)
x 4
1              x4
ln(     )e
x4
1                                                 cos( x  3)
1
lim 2      and lim ln(
x4
)e                           x4
  so the Indeterminate Limit Rule can be
x 4              x4     x4
applied on the limit to receive:
1
x 4
ln( 2          )
lim                              cos( x 3)
x 4
1                       x 4
ln(ln( )e      )
x4
Applying the product logarithm rule in the denominator results in:
1
x4
ln( 2            )
lim                                           cos( x 3)
x 4
1
ln(ln(    ))  ln( e x 4 )
x4
Applying the power logarithm rule on the numerator and denominator produces:
1
ln( 2)
lim          x4
x 4         1        cos(x  3)
ln(ln(      )) 
x4           x4
Multiplying the numerator and denominator by ( x  4) results in the following:
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ln( 2)
lim
x 4                  1
( x  4) ln(ln(      ))  cos(x  3)
x4
1
To get the limiting value of, lim ( x  4) ln(ln(                      )) , first let y = x – 4. Then, note
x 4                  x4
1                                                                1
that lim ( x  4) ln(ln(               )) can be expressed as lim ( x  4) ln(ln(                    )) since if x
x 4                   x4                                    ( x  4 ) 0              x4
approaches 4, x – 4 will approach 4 – 4 or 0. Next, substitute all occurrences of (x –
1                                1
4) with y to obtain a limit of lim y ln(ln( )) . Once lim y ln(ln( )) is expressed as
y0            y                 y0           y
1
ln(ln( ))
y                                                                                    1
lim         1
, L’Hopital’s Rule can be used on the limit since lim ln(ln( ))   and
y 0      y                                                                               y 0      y
1
(        )( y )( y 2 )
1                   1
ln(ln( ))                ln( )
1                            y                  y
lim y   so lim                  1
= lim                 2
. Cancelling the  y 2 term
y 0                  y 0      y             y 0         y
y
results in: lim                or 0. lim cos(x  3) = cos(1)
y0     1              x 4
ln( )
y
Applying the sum rule of limits on the denominator of
ln( 2)
lim                                                   results in the limiting value of
x 4                      1
( x  4) ln(ln(           ))  cos(x  3)
x4
1
lim ( x  4) ln(ln(            ))  lim cos(x  3) or 0 + cos(1), which is equivalent to cos(1).
x 4                  x4           x 4

ln( 2)
Using the quotient rule of limits on lim                                                            results in:
x 4                         1
( x  4) ln(ln(            ))  cos(x  3)
x4
ln( 2)
 1.282887
cos(1)
1                                          1
x 4                                        x4
ln( 2          )                                2
Therefore since lim                            cos( x 3)
>1, lim                   cos( x 3)
=  by the
x 4                                            x 4
1                    x 4
1           x4
ln(ln(     )e                           )              ln(     )e
x4                                                 x4
Indeterminate Limit Rule.

II. The Extended Indeterminate Limit Rule

Sometimes the Indeterminate Limit Rule needs to be used recursively. In these
cases, the Extended Indeterminate Limit Rule becomes useful.
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Corollary I:
Let ln n (h( x)) be defined as the nth natural logarithm of some function, h(x) , where n is
constrained to being a natural number. Defined recursively,
ln n (h( x)) = ln(ln n1 (h( x))) for all n greater than 1 and ln(h( x)) for n = 1. If lim f ( x)  
xa

and lim g ( x)   , where a is a finite number or a representation of the end behavior of
x a
the function (a =   ), then:

ln n ( f ( x))                f ( x)
1) If lim        n
 1, then lim         0
x a ln ( g ( x))             x a g ( x)

ln n ( f ( x))                f ( x)
2) If lim        n
 1 , then lim        
x a ln ( g ( x))              xa g ( x)

ln n ( f ( x))                f ( x)
3) If lim        n
  , then lim        
x a ln ( g ( x))              xa g ( x)

Proof:

The inductive proof of this rule is as follows:

Basis Step (n = 1): The Extended Indeterminate Limit Rule for n = 1 is identical to the
Indeterminate Limit Rule. Therefore, the proof of this corollary for n = 1 is identical to
the proof of the Indeterminate Limit Rule.

Inductive Step (If true for n, true for n+1): Let h(x) = ln n ( f ( x )) and j (x) = ln n ( g ( x)) .
By these assignments to h(x) and j (x) , ln(h( x)) = ln(ln n ( f ( x)))  ln n 1 ( f ( x)) and
ln( j ( x)) = ln(ln n ( g ( x)))  ln n 1 ( g ( x)) . Therefore proving the corollary true for n+1 is
h( x )
the same as proving the Indeterminate Limit true for lim                        . The functions, h(x)
x a j ( x )

and j (x) satisfy the condition that their limits as x approaches a are infinity. This
conclusion can be arrived at by realizing that the natural logarithm of a function will have
vertical asymptotes in the same locations that the original function has them, and the
natural logarithm of a function with positive unbounded end behavior will also have
positive unbounded end behavior. Therefore, if a finite number cannot be found for a
function’s limiting value at some point, a finite number will not be obtained for the limit
of the natural logarithm of the function at that point. This fact is also true of end
behavior. If a function has no finite limiting value for its end behavior, it’ll have no finite
limiting value for the end behavior of the natural logarithm of the function.
Consequently, taking the nth natural logarithm of a function, where n is a natural number,
will not make infinite limits finite. Therefore the Indeterminate Limit Rule can be used
h( x)
for lim           assuming lim f ( x)   and lim g ( x)   , where a is a finite number or a
x a j ( x )              xa                      x a
14

ln( h( x))
representation of the end behavior of the function (a =   ). If lim                                      1 , then
x  a ln( j ( x ))

h( x )                     ln( h( x))                  ln( h( x))                      h( x )
lim          0 , and if lim                  1 or lim                    , then lim               by the
xa j ( x)               x  a ln( j ( x ))          x  a ln( j ( x ))              x a j ( x)

Indeterminate Limit Rule. By the assumption that the Extended Indeterminate Limit
h( x )              ln n ( f ( x))                     f ( x)
Rule is true for n, if lim              0 ( lim n                  0 ), then lim             0 , and if
xa j ( x)           x a ln ( g ( x))                  x a g ( x)

h( x )               ln n ( f ( x))                     f ( x)
lim             ( lim n                    ), then lim                . These conclusions follow
x a j ( x)           x a ln ( g ( x))                 xa g ( x)

directly from the corollary and the fact that 0 is less than 1. Therefore, by the chain rule,
ln( h( x))               f ( x)                     ln( h( x))                ln( h( x))
if lim                  1 , lim            0 , and if lim                 1 or lim                ,
x  a ln( j ( x ))       x a g ( x)                x  a ln( j ( x ))        x  a ln( j ( x ))

f ( x)
lim            .
xa g ( x)

 By induction, the three parts of the Extended Indeterminate Limit Rule hold true.

Example II:
x
3e x x (ln x) x
lim          x
x 
e3

lim 3e x x (ln x) x   and lim e 3   so the Extended Indeterminate Limit Rule can
x                               x

x                               x 

be used in this situation. Applying the Rule for n = 2 on the limit results in:
x
ln 2 (3e x x (ln x) x )
lim                 x
x 
ln 2 (e 3 )
Applying the product logarithm rule on the numerator produces the following:
ln 2 (3e )  ln 2 ( x x )  ln 2 ((ln x) x )
x

lim                            x
x 
ln 2 (e 3 )
Expanding the ln 2 ( f ( x )) ’s results in:
ln(ln((3e )))  ln(ln( x x ))  ln(ln((ln x) x ))
x

lim                                x
x 
ln(ln(e 3 ))
Using the power logarithm rule on both the numerator and denominator provides the
following results:
ln(e x ln(3))  ln( x ln( x))  ln( x ln(ln( x)))
lim
x                       ln(3 x )
Using the power logarithm rule on the leading term of the denominator results in the
following:
ln( e x ln(3))  ln( x ln( x))  ln( x ln(ln( x)))
lim
x                       x ln(3)
15

Applying the product logarithm rule on the above limit, and simplifying ln( e x ) to x
results in the following:
x  ln(ln( 3))  ln( x)  ln(ln( x))  ln( x)  ln(ln(ln( x)))
lim
x                              x ln(3)
Since lim ( x  ln(ln(3))  ln( x)  ln(ln( x))  ln( x)  ln(ln(ln( x))))   and
x

lim x ln(3)   , L’Hôpital’s Rule can be applied on the limit to receive:
x 

1     1         1            1
(1                                      )
x x ln( x) x x ln( x) ln(ln( x))
lim
x                     ln(3)
1                1                                 1
Since lim 1  1 , lim  0 , lim                   0 , and lim                           0 , the sum
x         x  x        x  x ln( x )             x  x ln( x ) ln(ln( x ))

1       1        1             1
rule of limits can be applied on the limit, lim (1                                                 ) to
x        x x ln( x) x x ln( x) ln(ln( x))
obtain the following:
1           1            1                   1
lim 1  lim  lim               lim  lim                             1 0  0  0  0  1
x     x  x  x  x ln( x )   x  x    x  x ln( x ) ln(ln( x ))

1         1       1              1
(1                                         )
x x ln( x) x x ln( x) ln(ln( x))
Using the quotient rule of limits on lim                                                            results
x                        ln(3)
in:
1    1         1             1
lim (1                                        )
x      x x ln( x) x x ln( x) ln(ln( x))                1
           .91024 .
lim (ln(3))                           ln(3)
x 
x                              x
ln 2 (3e x x (ln x) x )              3e x x (ln x) x
Therefore since lim                      x
<1, lim               x
= 0 by the Extended
x                                  x 
ln 2 (e 3 )                                   e3
Indeterminate Limit Rule.

III. The Indeterminate Oblique Asymptote Limit Rules (Sum & Difference Rules)

Corollary II:

g ( x)                                        j ( x)
If lim f ( x)   ,  1  lim               1 , lim h( x)   , and  1  lim           1 (a can be a
xa                        x a f ( x)         x a                     x  a h( x )

finite number or it can represent the end behavior of the function, a =   ), then:
ln( f ( x))                    f ( x)  g ( x)
1) If lim                 1 , then lim                      0
x  a ln( h( x ))              x  a h( x )  j ( x )

ln( f ( x))                    f ( x)  g ( x)
2) If lim                 1 , then lim                      
x  a ln( h( x ))              x  a h( x )  j ( x )
16

ln( f ( x))                  f ( x)  g ( x)
3) If lim                   , then lim                    
x  a ln( h( x ))            x  a h( x )  j ( x )

Proof:

ln( f ( x))
Let L = lim                      . Factoring out f(x) from the numerator and h(x) from the
x a ln( h( x ))

g ( x)
f ( x)(1              )
f ( x)  g ( x)                                      f ( x)                     g ( x)
denominator in lim                                   results in lim                            . Since lim              and
x  a h( x )  j ( x )                 xa                 j ( x)                x a f ( x )
h( x)(1              )
h( x )
j ( x)                                                                                                  g ( x)
lim              are both finite numbers contained in the interval (-1, 1], lim (1                                    ) and
x a h ( x )                                                                                       xa         f ( x)
j ( x)
lim (1              ) will both be positive, finite numbers by the sum rule of limits. Therefore,
x a        h( x )
g ( x)
(1            )
f ( x)
lim                      is a positive finite number by the quotient rule of limits.
xa           j ( x)
(1            )
h( x )
g ( x)
(1            )
f ( x)                       f ( x)
1) L < 1: Let M = lim                                 . Let N = lim              . Since L is less than 1, N = 0 by
xa          j ( x)                 x a h ( x )
(1           )
h( x )
g ( x)
f ( x)(1           )
f ( x)
the Indeterminate Limit Rule. Since M is also finite, lim                                                      can be
xa                j ( x)
h( x)(1           )
h( x )
g ( x)
(1           )
f ( x)           f ( x)
expressed as lim                            lim            or MN by the product rule of limits. Since N = 0
xa         j ( x ) x  a h( x )
(1           )
h( x )
g ( x)
f ( x)(1            )
f ( x)
and M is finite, MN = 0. Therefore, lim                                                = 0. By distributing f(x) and
xa              j ( x)
h( x)(1            )
h( x )
h(x) in the numerator and denominator, respectively, we receive the desired results of
f ( x)  g ( x)
lim                         0.
x  a h( x )  j ( x )
17

ln( f ( x))                  ln( f ( x))             f ( x)
2,3) L > 1 or L =  : Since lim                            1 or lim                     , lim            by the
x  a ln( h( x ))            x  a ln( h( x ))       x  a h( x )

Indeterminate Limit Rule. In this case, the product rule of limits wouldn’t apply to
g ( x)
f ( x)(1            )
f ( x)
lim                           in the technical sense because the limit as x approaches a of one of
xa                j ( x)
h( x)(1            )
h( x )
the factors is not finite. However, a limit where one of its factors grows without bound in
the positive direction whereas the only other one approaches a positive constant will be
 . (For a formal proof of this statement, see the section entitled Aux. Theorem I at the
g ( x)
f ( x)(1           )
f ( x)
end of this paper.) Therefore, lim                                        . By distributing f(x) in the
xa               j ( x)
h( x)(1           )
h( x )
numerator and h(x) in the denominator we receive the desired results of
f ( x)  g ( x)
lim                      .
x  a h( x )  j ( x )

 The three parts of the Indeterminate Oblique Asymptote Limit Rules hold true.

Example III:
xx  ex
lim x
x  e  x ln( x )

 ex              e
lim      x
 lim ( ) x  0 . Since lim x ln( x )   and lim e x   , the Indeterminate
x  x            x  x                    x              x 
ln( x )
x
Limit Rule can be applied on lim x . Applying the Indeterminate Limit Rule on
x  e
ln( x )
x
lim x produces the following:
x  e

ln( x ln( x ) )
lim
x  ln( e x )

Applying the logarithmic power rule on the numerator and denominator results in the
following:
ln( x) ln( x)           (ln( x)) 2
lim                     lim
x     x ln( e)        x      x
Since lim (ln( x))2   and lim x   , L’Hôpital’s Rule can be applied on the limit to
x                        x 
receive:
18

2 ln( x)
(        )
x              2 ln( x)
lim             = lim
x     1         x     x
Since lim 2 ln( x)   and lim x   , L’Hôpital’s Rule can be applied on the limit to
x                     x 
receive:
2
( )
2
lim x  lim  0
x  1  x  x

ln( x ln( x ) )         x ln( x )
Therefore since lim                 <1, lim x = 0 by the Indeterminate Limit Rule.
x  ln( e x )          x  e

 ex                     x ln( x )
Since lim x   ,  1  lim x  1 , lim e   , and  1  lim x  1 , the
x                                   x
x                   x  x          x        x  e

xx  ex
Indeterminate Oblique Asymptote Limit Rules can be applied on lim x                 to
x  e  x ln( x )

receive:
ln( x x )
lim
x  ln( e x )

Applying the power logarithm rule on the numerator and denominator and canceling
the x’s results in the following:
x ln( x)         ln( x)
lim             lim          lim ln( x)  
x  x ln( e)    x    1       x 

ln( x x )             xx  ex
Therefore since lim            , lim x                by the Indeterminate Oblique
x  ln( e x )       x  e  x ln( x )

Asymptote Limit Rules.

IV. The Indeterminate Product Limit Rules
Corollary III:

f ( x)
If lim f ( x)   , lim ( f ( x)(g ( x)) y )   , lim                 , lim h( x)   ,
xa                  x a                      xa    ( g ( x)) y       x a

h( x )
lim (h( x)( j ( x)) y )   , and lim                  for all positive finite y (a can be a finite
x a                              x  a ( j ( x )) y

number or it can represent the end behavior of the function, a =   ), then:
ln( f ( x))                    f ( x) g ( x)
1) If lim                1 , then lim                     0
x  a ln( h( x ))             x  a h( x ) j ( x )

ln( f ( x))                    f ( x) g ( x)
2) If lim                1 , then lim                     
x  a ln( h( x ))             x  a h( x ) j ( x )

ln( f ( x))                      f ( x) g ( x)
3) If lim                 , then lim                      
x  a ln( h( x ))               x  a h( x ) j ( x )
19

Proof:
f ( x)
If a =  , then for any positive finite y, N  0 : x  N ,                     1 , f ( x)  1 , and
( g ( x)) y
f ( x)
f ( x)( g ( x)) y  1 . If a = –  , , then for any positive finite y, N  0 : x  N ,                  1,
( g ( x)) y
f ( x)  1 , and f ( x)( g ( x)) y  1 . Otherwise, if a is finite, , then for any positive finite y,
f ( x)
there exists a positive delta such that for all x, where a    x  a   ,                      1,
( g ( x )) y
f ( x)  1 , and f ( x)( g ( x)) y  1 . These conclusions follow directly from the formal
definition of limits. (Since an expression with the limiting value of infinity means that
the expression is greater than any finite number over some domain, a domain for which
the expression is greater than 1 can be found.) To find an N or a  that simultaneously
f ( x)
satisfies                 1 , f ( x)  1 , and f ( x)( g ( x)) y  1 , select the smaller  or the N with
( g ( x )) y
the greater absolute value that satisfies both conditions, if a is finite or infinite,
f ( x)
respectively. For all assignments to a, there is some domain for which                                  1,
( g ( x )) y
f ( x)  1 , and f ( x)( g ( x)) y  1 . Some algebra can be performed on the first listed
inequality:

Multiplying the inequality through by ( g ( x)) y , the following is obtained (The
inequality sign does not change because ( g ( x)) y is always positive over the restricted
f ( x)
domain. ( g ( x)) y has this attribute since f ( x)  1 and               1 over the specified
( g ( x )) y
domain. A positive fraction with a positive numerator must have a positive
denominator for the fraction to be positive.):

f ( x)  ( g ( x)) y

Taking the natural logarithm of both sides results in (The inequality sign does not

change because the natural logarithmic function is an increasing function over its

entire domain.):

ln( f ( x))  ln(( g ( x)) y )

Performing the power logarithm rule on the right side of the inequality produces the

following:
20

ln( f ( x))  y ln( g ( x))

Dividing the inequality through by y ln( f ( x)) results in (The inequality sign does not

change because y and ln( f ( x)) are always positive over the restricted domain forcing

their product to also be positive.):

1 ln( g ( x))

y ln( f ( x))

f ( x)
Since the domain for which                         1 and f ( x)  1 are true is the domain
( g ( x )) y

1 ln( g ( x))
determined by the limit as x approaches a, and                         is an algebraic
y ln( f ( x))

f ( x)
manipulation of                       1 for f ( x)  1 , the limit as x approaches a can be applied
( g ( x )) y

on the inequality to receive:

1         ln( g ( x))
lim       lim
x a   y   x  a ln( f ( x ))

Since y is unaffected by the limit, the above inequality can be simplified to the

following:

1       ln( g ( x))
 lim
y xa ln( f ( x))

1
Since y was assigned to represent any positive finite number,               can be made to get
y

1
as close to zero as possible by selecting a large enough y value. (             can obtain any
y

1
positive z value by assigning y to           .) To represent the fact that the left-hand side of
z
21

the inequality can be made to be smaller than any positive number, the inequality can

be expressed as the following:

ln( g ( x))
0  lim                (2)
x a ln( f ( x ))

Some algebra can also be performed on the third listed inequality ( f ( x)( g ( x)) y  1 ):

Dividing the inequality through by f (x) results in (The inequality sign does not
change because f (x) is always positive over the restricted domain.)

( g ( x)) y  ( f ( x)) 1

Taking the natural logarithm of both sides results in (The inequality sign does not

change because the natural logarithmic function is an increasing function over its

entire domain.):

ln(( g ( x)) y )  ln(( f ( x)) 1 )

Performing the power logarithm rule on both sides of the inequality produces the

following:

y ln( g ( x))   ln( f ( x))

Dividing the inequality through by y ln( f ( x)) results in (The inequality sign does not

change because y and ln( f ( x)) are always positive over the restricted domain forcing

their product to also be positive.):

ln( g ( x))    1

ln( f ( x))    y

Since the domain for which f ( x)( g ( x)) y  1 and f ( x)  1 are true is the domain

ln( g ( x))    1
determined by the limit as x approaches a, and                   is an algebraic
ln( f ( x))    y
22

manipulation of f ( x)( g ( x)) y  1 for f ( x)  1 , the limit as x approaches a can be

applied on the inequality to receive:

ln( g ( x))          1
lim                 lim ( )
x  a ln( f ( x ))   xa   y

Since y is unaffected by the limit, the above inequality can be simplified to the

following:

ln( g ( x))    1
lim               
xa   ln( f ( x))    y

1
Since y was assigned to represent any positive finite number,          can be made to get
y

1
as close to zero as possible by selecting a large enough y value. (       can obtain any
y

1
negative z value by assigning y to      .) To represent the fact that the right-hand side
z

of the inequality can be made to be larger than any negative number, the inequality

can be expressed as the following:

ln( g ( x))
lim                0 (3)
xa   ln( f ( x))

Putting inequalities (2) and (3) together results in the following:

ln( g ( x))
0  lim                0
x  a ln( f ( x ))

ln( g ( x))
By the Squeeze Law, we obtain the results: lim                     0.
x  a ln( f ( x ))

By substituting h(x) for f (x) and j (x) for g (x) in the above proof, the following is

obtained:
23

ln( j ( x))
lim               0
x  a ln( h( x ))

f ( x) g ( x)               ln( f ( x) g ( x))
By the Indeterminate Limit Rule, lim                                 0 if lim                        1 . Let L =
xa   h( x ) j ( x )        x  a ln( h( x ) j ( x ))

ln( f ( x))
lim                . Factoring out ln( f ( x)) from the numerator and ln(h( x)) from the
x a  ln( h( x))
ln( g ( x ))
ln( f ( x))(1                  )
ln( f ( x))  ln( g ( x))                                        ln( f ( x))
denominator in lim                                       results in lim                                       . Since
x  a ln( h( x ))  ln( j ( x ))                xa                    ln( j ( x ))
ln( h( x))(1                 )
ln( h( x ))
ln( g ( x))               ln( j ( x))                             ln( g ( x))                      ln( j ( x))
lim                 and lim                 are both 0, lim (1                     ) and lim (1                    ) will
x a ln( f ( x ))        x a ln( h ( x ))                  xa        ln( f ( x))           xa        ln( h( x))
ln( g ( x))
(1               )
ln( f ( x))
both be 1 by the sum rule of limits. Consequently, lim                                               is also 1 by the
xa        ln( j ( x))
(1               )
ln( h( x))
ln( g ( x))
(1               )
ln( f ( x))
quotient rule of limits. Let M = lim                                     . Since L represents the ratio of the
xa         ln( j ( x))
(1               )
ln( h( x))
natural logarithms of two functions whose limiting values are positive infinity, the ratio
can’t be negative. Therefore, L must be positive. Therefore, L is either finite or L =  .
ln( g ( x ))
ln( f ( x))(1                 )
ln( f ( x))
Since M is finite, lim                                          can be expressed as
xa                   ln( j ( x ))
ln( h( x))(1                )
ln( h( x ))
ln( g ( x))
(1               )
ln( f ( x))          ln( f ( x))
lim                       lim                or ML by the product rule of limits if L is finite. Since M
xa        ln( j ( x)) x a ln( h( x))
(1               )
ln( h( x))
ln( g ( x ))
ln( f ( x))(1                 )
ln( f ( x))
= 1, the limit simplifies to (1)L = L. If L =  , lim                                                      = L by Aux.
xa                    ln( j ( x ))
ln( h( x))(1                )
ln( h( x ))
24

ln( g ( x ))
ln( f ( x))(1             )
ln( f ( x))
Theorem I. Therefore, either way, lim                                                = L. By distributing
xa                ln( j ( x ))
ln( h( x))(1               )
ln( h( x ))
ln( f ( x)) and ln(h( x)) in the numerator and denominator, respectively, we receive the
ln( f ( x))  ln( g ( x))
results of lim                                 = L, which by the product logarithm rule can be
x  a ln( h( x ))  ln( j ( x ))

ln( f ( x) g ( x))                          ln( f ( x) g ( x))          ln( f ( x))
rewritten as lim                           = L. Therefore, lim                         = lim               .
x a ln( h( x ) j ( x ))                    x a ln( h( x ) j ( x ))    x a ln( h( x ))

lim ( f ( x) g ( x))   and lim (h( x) j ( x))   by the prerequisites that
x a                              x a

lim ( f ( x)(g ( x)) )   and lim (h( x)( j ( x)) y )   for y = 1. The rest of the proof follows
y
x a                                     x a
directly from the Indeterminate Limit Rule:

ln( f ( x) g ( x))             ln( f ( x))                ln( f ( x))
1) From the conclusion that lim                                      = lim                  , if lim                  1 , then
x a  ln( h( x) j ( x))        x a ln( h( x ))          x  a ln( h( x ))

ln( f ( x) g ( x))                 ln( f ( x) g ( x))                        f ( x) g ( x)
lim                         1 . If lim                          1 , then lim                        0 by the
x  a ln( h( x ) j ( x ))          x  a ln( h( x ) j ( x ))                 x  a h( x ) j ( x )

ln( f ( x))                f ( x) g ( x)
Indeterminate Limit Rule. Therefore, if lim                                     1 , lim                      0 by the chain
x  a ln( h( x ))         x  a h( x ) j ( x )

rule.
ln( f ( x) g ( x))             ln( f ( x))                ln( f ( x))
2) From the conclusion that lim                                      = lim                  , if lim                  1 , then
x a ln( h( x ) j ( x ))       x a ln( h( x ))          x  a ln( h( x ))

ln( f ( x) g ( x))                 ln( f ( x) g ( x))                        f ( x) g ( x)
lim                         1 . If lim                          1 , then lim                         by the
x  a ln( h( x ) j ( x ))          x  a ln( h( x ) j ( x ))                 x  a h( x ) j ( x )

ln( f ( x))                f ( x) g ( x)
Indeterminate Limit Rule. Therefore, if lim                                     1 , lim                       by the
x  a ln( h( x ))         x  a h( x ) j ( x )

chain rule.
ln( f ( x) g ( x))             ln( f ( x))                ln( f ( x))
3) From the conclusion that lim                                      = lim                  , if lim                   , then
x a ln( h( x ) j ( x ))       x a ln( h( x ))          x  a ln( h( x ))

ln( f ( x) g ( x))                   ln( f ( x) g ( x))                         f ( x) g ( x)
lim                          . If lim                             , then lim                          by the
x  a ln( h( x ) j ( x ))            x  a ln( h( x ) j ( x ))                  x  a h( x ) j ( x )

ln( f ( x))                  f ( x) g ( x)
Indeterminate Limit Rule. Therefore, if lim                                      , lim                        by the
x  a ln( h( x ))           x  a h( x ) j ( x )

chain rule.

 The three parts of the Indeterminate Product Limit Rules hold true.

Example IV:
25

2.8 x
lim
x  10 x 50 e x ln(10 x)

Since lim e x   and lim (10x 50 ln(10x)) y   for all positive y, the Indeterminate
x                   x 

ex
Limit Rule can be applied on lim                               to receive the following:
x  (10 x 50 ln(10 x)) y

ln(e x )
lim
x  ln((10 x 50 ln(10 x)) y )

Applying the power logarithm rule on the numerator and denominator results in:
x
lim             50
x  y ln((10 x    ln(10 x)))
Using the product logarithm rule on the denominator produces:
x
lim
x  y (ln(10 )  ln( x )  ln(ln(10 x )))
50

Since lim x   and lim ( y(ln(10)  ln( x 50 )  ln(ln(10 x))))   , L’Hôpital’s Rule can
x              x 
be applied on the above limit to receive:
1
lim
x     50       1
y(               )
x x ln(10 x)
Distributing the y in the denominator results in:
1
lim
x  50         1
y            y
x     x ln(10 x)
Multiplying the numerator and denominator by lim x ln(10x) results in:
x 

x ln(10 x)
lim
x  50 y ln(10 x )  y

Since lim x ln(10x)   and lim (50 y ln(10x)  y)   , L’Hôpital’s Rule can be
x                          x 
applied on the above limit to receive:
ln(10 x)  1
lim
x       50 y
(      )
x
Multiplying the numerator and denominator by x results in:
x ln(10 x)  x
lim                 
x        50 y
ln(e x )                              ex
Because lim                              =  , lim                       =  by the Indeterminate
x  ln((10 x 50 ln(10 x)) y )       x  (10 x 50 ln(10 x)) y

2.8 x
Limit Rule. Since lim 2.8 x   , lim ((2.8 x )(1) y )   , lim             , lim e x   ,
x                  x                   x  1 y        x 
26

ex
lim (e x (10 x 50 ln(10 x)) y )   , and lim                      =  for all positive finite y,
x                                      x  (10 x 50 ln(10 x)) y

2.8 x
the Indeterminate Product Limit Rules can be applied on lim                                   to
x  10 x 50 e x ln(10 x)

receive the following:
ln( 2.8 x )
lim
x  ln( e x )

Applying the power logarithm rule on the numerator and denominator results in the
following:
x ln( 2.8)
lim               ln( 2.8)  1.02962
x  x ln( e)

ln( 2.8 x )                   2.8 x
Therefore since lim                 > 1, lim                         by the Indeterminate Product
x  ln( e x )       x  10 x 50 e x ln(10 x)

Limit Rules.

V. The Indeterminate Product Limit Rules: A Closer Look
Although dominant terms can be found within sums, there is no current method of
f ( x)
finding dominant terms within products. If some function, f(x), satisfies lim        ,
xa g ( x)

then f ( x)  g ( x) behaves like f(x) as x approaches a. For example, x 3  10x 2 behaves
like x 3 for very large x. Because only dominant terms need to be considered when
dealing with limits of sums, it can be said that the lesser terms don’t affect the limit. This
dominant term idea already exists, and it is this dominant term consideration that forms
the basis for the Indeterminate Oblique Asymptote Limit Rules. However, what is the
dominant term in a product such as 10 xe x ? Is there such a term? When will there be
dominant terms? What defines dominance within products? What are the implications of
dominance within products?
The answers to these questions are encompassed in the Indeterminate Product
Limit Rules. A sum mimics the behavior of its most dominant term when approaching
the specific value or extreme where it dominates. A product has no such behavior.
However, there are often terms within a product which can determine the limiting value
of the entire product in certain situations. Specifically, these cases occur when the limit
p( x)
is in the form, lim          , where p(x) and q(x) are functions or products of functions
x a q ( x )

such that lim p( x)   and lim q( x)   . They can also occur when the limit is in the
x a                x a

p( x)  r ( x)
form, lim                   where p(x) , q(x) , r (x) , and s(x) are functions or products of
xa   q( x)  s ( x)
27

functions if the limit satisfies the prerequisites for the usage of the Indeterminate Oblique
p( x)  r ( x)
Asymptote Limit Rules. The reason lim                       is a valid form for dominance
xa q( x)  s ( x)

examination so long as the Indeterminate Oblique Asymptote Limit Rules can be applied
to the fraction is that an application of the Rules reduces the numerator and denominator
to single terms of products satisfying the conditions, lim p( x)   or lim r ( x)   and
x a              x a

lim q( x)   or lim s( x)   . Therefore, the limit can be reduced to a form that can be
x a                x a

handled by the Indeterminate Product Limit Rules (the first mentioned form) prior to
p( x)  r ( x)
determining dominace. Therefore, lim                 can be reduced to a form that can be
xa q( x)  s ( x)

handled by the Indeterminate Product Limit Rules following an application of the
Indeterminate Oblique Asymptote Limit Rules since Corollary I reduces the numerator
and denominator to a single term of products. If a dominant term within a product is
considered to be one that permits other terms within the product to be ignored without
changing the limit in these forms, then according to the Indeterminate Product Limit
Rules, a function, f (x) , dominates another function, g (x) , in a limit where x approaches
f ( x)
a if lim ( f ( x)(g ( x)) y )   and lim                 for all positive finite y. By showing that
x a                          xa    ( g ( x)) y
the Indeterminate Limit Rule produces the same results for a limit with all its terms as it
does for just the most dominant ones in the numerator and denominator, the
Indeterminate Product Limit Rules Corollary would seem to suggest that finding the
limiting value of a fraction with a single product in the numerator and denominator is as
simple as finding the limit of the dominant terms in each location. Although this
generalization is normally true, there are some exceptions to this rule.
The exceptions to this rule of ignoring lesser terms in products can occur and will
only occur when the ratio produced by using the Indeterminate Limit Rule results in 1.
The conclusion that if the Indeterminate Limit Rule yields the same number for two
different limits, then the limiting values are equivalent will always be true so long as the
test produces conclusive results. However, when the test results in a ratio of 1, the
Indeterminate Limit Rule says nothing about the limit. Therefore, a limit where the
limiting value of the ratio of the natural logarithms of the numerator to the denominator is
1 can be equivalent to any finite number or it can be infinite.
Because the Indeterminate Limit Rule only uses the most dominant terms in the
numerator and denominator in determination of the limit, the test will result in a ratio of 1
and yield inconclusive results when both parts of the fraction could contain the same
dominant term. For example, the ratio resulting from the application of the Indeterminate
28

x2 x ln( x)            2x
Limit Rule to lim                  and lim x will both be 1 because the dominant terms in
x  10  2 x ln( x)     x 2

the numerator and denominator could both be 2 x . (However, this example does not
counter the Indeterminate Product Limit Rules because although
x2 x ln( x)          2x         ln( x2 x ln( x))                                   ln( 2 x )
lim                   lim x , lim                      will still be equivalent to lim            or
x  10  2 x ln( x)   x 2   x  ln(10  2 x ln( x))                             x  ln( 2 x )

simply 1.) Sometimes, the same dominant terms could appear in both the numerator and
x!
denominator but will take different forms. For example, lim x , could have the same
x  x

dominant terms in both the numerator and denominator because lim x! can be rewritten
x 
x
x                                                                           x!
as lim ((     x
) 2x ) by Sterling’s Formula. If the numerator and denominator of lim x
x      e                                                                      x  x

xx                     x!
are multiplied by e and x! is replaced with (( x ) 2x ) , then lim x can be expressed
x

e                 x  x

x x 2x
as lim           . Because the numerator and denominator could both have the same
x    x xe x
x x 2x
dominant term of x x , the Indeterminate Limit Rule will result in 1 for lim                   and
x    x xe x
x!
the equivalent limit of lim    . This same example can illustrate another potential way
x  x x

x!
dominant terms can be disguised in a more conspicuous manner. If once lim x was
x  x

x x 2x
rewritten as lim           , we let x x and x x e x be the dominant terms of the numerator
x    x xe x
and denominator respectively, then the dominant terms are not the same. However, this
selection of dominant terms will not change the fact that applying the Indeterminate
x x 2x
Limit Rule to the limit of lim           will still yield the inconclusive ratio of 1.
x    x xe x
Therefore, it would seem that duplicate dominant terms would not be the only time when
the Indeterminate Limit Rule would result in inconclusive results. However, that’s why
“could” is the word that’s used in this paragraph when describing dominant term
selection. From the definition of dominance, we can find an infinite number of dominant
x x 2x                           x x 2x
terms by rewriting limits, (e.g. lim             can be expressed as lim 2 x  2 x by
x    x xe x                     x  x x   e
x x ln( x) 2x                   ln( x )
expanding the x x term or lim           x x
by multiplying by         .) However, when
x    x e ln( x)                     ln( x )
29

the Indeterminate Limit Rule produces inconclusive results, a dominant term that appears
in both the numerator and denominator could be found either visibly or by rewriting a
function that appears in the numerator and/or denominator, without creating dominant
ln( x)   2x
terms (e.g. without multiplying lim            by x ), as an equivalent one.
x    x      2
The dominant term concept incorporated in the Indeterminate Product Limit Rules
can still be used in these situations where the dominant terms could be the same for the
numerator and denominator if a preliminary step is taken. This step is the systematic
cancellation of dominant terms (either shown or concealed such as in lim x! , which has
x 
x
x
the dominant term of x x when rewritten as lim ((   ) 2x ) ) until the dominant terms in
ex
x 

the numerator and denominator, f (x) and g (x) , respectively have the property that
f ( x)
either lim                 0 for some value of y falling in the open interval (0, l) or
x  a ( g ( x )) r

f ( x)
lim                  for some value of y falling in the open interval (1,  ) . (This property,
x  a ( g ( x )) r

which may seem arbitrary, ensures that one of the dominant terms is not only not a
duplicate of the other one, but also not a duplicate of a factor contained within the other
dominant term if the contained factor could dominate the rest of the product. For
example, dominant terms of 2 x and 10 x2 x in a limit where x approaches infinity would
not satisfy this property because 2 x is a duplicate of the factor, 2 x , in 10 x2 x and 2 x
dominates 10 x . However, dominant terms of x 3 and x 2 in a limit where x approaches
infinity would satisfy this property because even though x 3 can be expressed as x  x 2 ,
x 2 does not dominate x in a product. The property also ensures that the limiting value of
the ratio of the natural logarithms of the dominant terms will not be 1, thus ensuring a
conclusive result from the Indeterminate Limit Rule.) For example, the dominant term
x2 x ln( x)
concept can still be used on lim                    . In this limit, the dominant terms of the
x  10  2 x ln( x)

x                                   2x
numerator and denominator are both 2 . Once these terms are eliminated ( lim x r
x  (2 )

will not satisfy one of the prerequisites since when y is less than 1, the limiting value is
infinity and when y is greater than 1, the limiting value is 0.), x becomes the dominant
term in the numerator and ln(x) becomes the dominant term in the denominator. Since
x
lim                  for any positive finite value of y, no further cancellations are needed,
x   ( ln( x )) y
30

x
and the limit can be simplified to lim             . If the x did not appear in the numerator, the
x   ln( x )

ln(x) would become the dominant term after cancellation in both the numerator and
denominator. If this were the case, the ln(x) would also have to be cancelled before
xx
using the dominant term concept. In a limit such as lim          where a duplication of
x   3 x x!

dominant terms in both the numerator and denominator is disguised, the limit should be
expressed as an equivalent one where the duplicate dominant terms are shown. Using
Sterling’s Formula and multiplying the numerator and denominator by e x expresses
xx                                        x xe x
lim          as the equivalent limit of lim x             . After canceling the duplicate
x   3 x x!                            x 
( x 2x )3 x
dominant terms of x x , e x and 3 x become the new dominant terms of the numerator and
ex                            1
denominator, respectively. Since lim x y  0 for any y greater than        (~ .9102), a
x  (3 )                          ln( 3)
f ( x)
y value less than 1 that satisfies the prerequisite that lim                  0 for some value of y
x  a ( g ( x )) r

falling in the open interval (0, l) can be found (e.g. .95). Therefore, the limiting value of
xx                                                    ex
lim x will be equivalent to the limiting value of lim x .
x   3 x!                                              x  3

The prerequisites for dominance as outlined by the Indeterminate Product Limit
Rules are in place in order to ensure that the lesser terms in the product do not affect the
results of imposing the Indeterminate Limit Rule on the limit. For this to happen, the
ratio of the natural logarithm of the lesser function to the natural logarithm of the
dominant one must have a limiting value of 0. While it is sufficient for a dominant term,
f ( x)
f(x), in a sum of f ( x)  g ( x) to have the property that lim           , a dominant product,
xa g ( x)

f ( x)
f(x), in a product of f ( x) g ( x) must have the property that lim                 for all
xa   ( g ( x)) y
positive finite y. Therefore x 2 dominates x in a sum, but not in a product. As was
mentioned in the introduction, the faster a function grows, the greater the growth
decrease will be from the function to its natural logarithm. Therefore, if f (x) is the
dominant term in f ( x) g ( x) , f (x) must significantly outgrow g (x) . (Note that the ratio
of the logarithms of x 2 to x is only 2 despite the fact that the ratio of the two functions is
x. When x becomes very large, so does the ratio of the two functions. However, the ratio
of the logarithms as x becomes very large is still just 2.) If lim f ( x)   and
xa
31

f ( x)                                        ln( g ( x))
lim              y
  for all positive finite y, lim               is ensured to not be positive.
x  a ( g ( x ))                                  x a ln( f ( x ))

Therefore, having f ( x) g ( x) instead of f (x) will not increase the ratio of the natural
logarithms of the numerator to the denominator if the product appears in the numerator
and will not decrease the ratio if it appears in the denominator. However, g (x) can still
decrease the numerator or denominator of the ratio produced by the Indeterminate Limit
Rule if f ( x) g ( x) appears in the numerator or denominator, respectively. Adding the
condition that lim ( f ( x)(g ( x)) y )   for f (x) to be the dominant term in f ( x) g ( x)
x a

ln( g ( x))
ensures that lim                 will not be negative. The cases where this added condition will
x a ln( f ( x ))

1
be necessary is when lim g ( x)  0 , and lim                would not be able to be dominated by
x a                  x a   g ( x)
f (x) . To illustrate the necessity of this second prerequisite, consider the limit,
2 x 2  x ln( x)
lim                   . If we let f (x) = 2 x ln( x) and g (x) = 2  x and ignore the condition that
x       10 x
lim ( f ( x)(g ( x)) y )   is necessary for dominance, then f (x) is a dominant term, and
x a

2 x ln( x)
the limit for which the Indeterminate Limit Rule is imposed on becomes lim                    .
x     10 x
Here, the ratio of the logarithms is infinity whereas the ratio of the logarithms of
2 x 2  x ln( x)
lim                   is zero. These resulting ratios would translate into the Indeterminate
x       10 x
Limit Rule producing the two very different limiting values of infinity and zero. In this
example, moving the reciprocal of 2  x to the denominator would result in a 2 x term
being in this location, and 2 x would not be dominated by 2 x ln( x) . Therefore the added
condition that lim ( f ( x)(g ( x)) y )   for f (x) to be the dominant term in the product,
x a

f ( x) g ( x) , is in place to make sure that g (x) does not approach very small numbers just
as fast or faster than f (x) approaches the reciprocal of these very small numbers as x
becomes very large.
It was mentioned earlier that the Indeterminate Limit Rule only uses the most
dominant terms in the numerator and denominator in its determination of the limit. This
property of the Indeterminate Limit Rule is the result of the logarithms’ attribute that a
logarithm of products can be expressed as the sum of the logarithms of the products. It is
also the result of the sums’ dominant term property. As was shown in the proof for
f ( x)
Corollary II, if some function, f(x), satisfies lim                  and lim ( f ( x)(g ( x)) y )  
x  a ( g ( x )) y         x a
32

ln( g ( x))
for all positive finite y with lim f ( x)   , then lim                   0 . Since
xa                    x  a ln( f ( x ))

ln( g ( x))
lim                 0 , ln( f ( x))  ln( g ( x)) will behave like ln( f ( x)) for x values approaching
x  a ln( f ( x ))

a because of the sum dominant term consideration. However, by the product limit rule of
logarithms, ln( f ( x) g ( x)) = ln( f ( x))  ln( g ( x)) . Therefore, ln( f ( x) g ( x)) will behave
like ln( f ( x)) for x values approaching a. Because the natural logarithm of the entire
product exhibits the same behavior as the natural logarithm of the most dominant term for
x values approaching a, the Indeterminate Limit Rule will yield the same ratio regardless
of whether the lesser terms are included or not. Therefore, the Indeterminate Limit Rule
ln( 2)                                   10 x 99 2 x
will produce the ratio,        , when the Rule is applied to lim x           ,
ln( 3)                               x  3 ln( x)

10 x 99 2 x ln( x)          2x
lim                      , or lim x (all of which have limits of 0 according to the results from
x          3x              x 3

ln( f ( x))
the Indeterminate Limit Rule), and the ratio, lim                    when the Rule is applied to
x a ln( h( x ))

f ( x) g ( x)
lim                   if f (x) and h(x) could be the dominant functions of the numerator and
x a   h( x ) j ( x )
denominator, respectively.
33

Conclusion
Although the Indeterminate Limit Rule can only be used for specific
indeterminate forms, it can be a useful tool with some important consequences. Limits

containing exponential functions in the indeterminate form, , can be easily solved

using the theorem. In addition, because the Indeterminate Limit Rule can be proven true,
a concept of dominance within products (and sometimes sums of products) can be
achieved. Through analysis of this Rule, we can also find out when a function can be
expressed in terms of a product of functions at a certain value or at certain extremes.

The Indeterminate Limit Rule makes solving limits in the indeterminate form, ,

th
easier. L’Hôpital’s Rule can easily handle unbounded functions with n derivatives (n is
any natural number) that approach finite numbers at the location where the function is
unbounded. For example, x 99 exhibits this property for its end behavior because the
function is unbounded as x becomes arbitrarily large, and the 100th derivative of the
function approaches 99! as x becomes arbitrarily large. ln(x) also exhibits this property
for its end behavior because its 1st derivative approaches 0 as x becomes arbitrarily large
whereas the function itself is unbounded as x attains increasingly large values. However,
most exponential functions will not exhibit this trait and will have nth derivatives that will
not have any finite limiting value at the locations where the function is unbounded. Since
the growth of the natural logarithm of an unbounded function will be significantly
smaller than the growth of the function at its unbounded location, the resulting ratio from
the Indeterminate Limit Rule and its corollaries will be able to be handled by L’Hôpital’s
Rule. Regardless of how fast the functions grow in the numerator and denominator, the
application of a finite number of iterations of the natural logarithm to the numerator and
denominator will make both respective locations contain only functions where the nth
derivatives (n is any natural number) will approach finite numbers at the location where
the function is positively unbounded (approaching arbitrarily high positive numbers).
Therefore, the application of the Extended Indeterminate Limit Rule can produce a ratio
that can be handled by L’Hôpital’s Rule.
By breaking products of functions into sums, the Indeterminate Limit Rule and its
corollaries change the behavior of products of functions into a function which only
reflects the behavior of the smallest dominant term*. Therefore the ratio that results from
the application of the Indeterminate Limit Rule to a fraction with products in both the
numerator and denominator will be equivalent to the ratio that would result from the
application of the Rule on a fraction with only the numerator’s smallest dominant term in
the numerator and the denominator’s smallest dominant term in the denominator.
Therefore, since the resulting ratio would be the same regardless of whether any or all of
34

the dominated terms are included or not, the limiting values of limits containing some or
all the dominated terms must be the same as a limit where the numerator and
denominator consist of only the numerator and denominator’s smallest dominant terms,
respectively so long as the ratio is not 1.
Because products do not exhibit the sole behavior of their most dominant terms,
the most we can expect a dominant term to do in a product is to solely determine the
limiting value in certain situations. As was mentioned in the previous section, these
p( x)
situations can occur when the limit is in the form, lim          , where p(x) and q(x) are
x a q ( x )

functions or products of functions such that lim p( x)   and lim q( x)   . Since as
x a                  x a

long as the Indeterminate Limit Rule does not result in the inconclusive ratio of 1,
ignoring any or all dominated terms within the products will not change the limiting
value of limits in these forms, this desired concept of dominance has been mostly
achieved. Since it was noted in the previous section that a ratio of 1 will occur when the
smallest dominant term in the numerator and denominator are the same, it can be said that
dominance will always be realized in these limit forms so long as the smallest dominant
term in the numerator and denominator don’t match. Because this situation can be
avoided by simple cancellation, this concept of dominance introduced by the
p( x)
Indeterminate Product Limit Rules can be applied to any limit in the form, lim             , so
x a q ( x )

long as p(x) and q(x) are functions or products of functions such that lim p( x)   and
x a

p( x)  r ( x)
lim q( x)   . Furthermore, if a limit in the form lim                    , where p(x) , q(x) ,
x a                                                  xa   q( x)  s ( x)
r (x) , and s(x) are functions or products of functions that meet the prerequisites for the
Indeterminate Oblique Asymptote Limit Rules to be applied to it, then the dominance
concept can be used for the limit in this form as well.
Since the Indeterminate Limit Rule has the unique property of only reflecting the
behavior of the smallest dominant terms for the numerator and denominator in the same
way (natural logarithms of the functions), the theorem can have useful applications. One
such application is finding a continuous function that mimics a positively unbounded
discrete function at the location where it is positively unbounded. For example, suppose
you wanted to find a continuous function that behaves like x! for very large x.
Furthermore, suppose Sterling’s Formula didn’t exist, and there was no known
continuous function that mimics the end behavior of x! . In this situation, the
Indeterminate Limit Rule can be used to help you find the answer in steps. It would be
xx
difficult to know that x! behaves like (( x ) 2x ) for very large x just by looking at the
e
outputs of the factorial function for large inputs. However, it isn’t as difficult to know
that the solution contains the term, x x . Then, after determining that the answer contained
35

x!
x x , it wouldn’t be too difficult to determine that          behaved like it contained the
xx
ex
reciprocal of an exponential function for very large x. If x! was multiplied by x , it
x
wouldn’t be very difficult to see that the new resulting function behaved like k x ,
where k is a constant, for very large x. The k could then be approximated after
ex
multiplying x! by x           . This illustrates the method that can be used by utilizing the
x x
Indeterminate Limit Rule. By taking the natural logarithm of x! for very large x values
and dividing by the natural logarithm of x x for these same very large x values, it can be
noted that the answer will be close to 1. From the observations made by the
Indeterminate Limit Rule in this report, we can conclude then that x x is probably the
smallest dominant term in the numerator of the continuous function that mimics x! ’s end
behavior. With this information, we can proceed to find the next smallest dominant term
(the term that would be the smallest dominant one if x x was not contained in the
continuous function that mimics x! ’s end behavior). To be able to find this next term, we
can divide x! by x x and examine the resulting values for very large x. Since the
resulting values approach 0, the reciprocal will have to be used in order to find the next
smallest dominant term. If we examine the ratio of the natural logarithms of the
x!                            xx
reciprocal of x , which is equivalent to           , to e x , we should see the ratio converge to 1
x                              x!
as x becomes very large. From the conclusion that the Indeterminate Limit Rule only
uses the smallest dominant terms, it would follow that the continuous function that
xx                                                                              xx
mimics ’s end behavior contains e x in its numerator. We can now divide                       by e x
x!                                                                             x!
xx              xx
to receive         . Since           approaches 0, the reciprocal will have to be used in order
x!e x           x!e x
xx    x!e x
to find the next smallest dominant term. Since the reciprocal of               , x appears to
x!e x    x
be approaching k x , where k is a constant, for very large x, we can then take the ratio of
x!e x
the natural logarithms of x to x to see if that is the case. Since the ratio
x
x!e x                 x!e x
approaches 1, we can conclude that it is and divide x by               x to get x         . It can
x                   x x
x!e x
then be noted that x           approaches 2.5066… as x becomes very large. Therefore, if we
x x
x!e x
let, k be this constant value, then dividing x           by k will result in 1 as x becomes very
x x
36

x!e x                        x!
large. From this information, we can conclude that lim                               1 or lim                1.
x         x              x       x
kx       x                   kx x
(      )
ex
x!
From lim                  1 , it can be determined that the end behavior of x! behaves like
x       x
kx x
(      )
ex
kx x x
for some constant, k, that is approximately 2.5066. As is illustrated from this
ex
process applied to the end behavior of x! , the Indeterminate Limit Rule allows an
unbounded product of functions to be decomposed into its parts. Consequently, if a
function has an infinite limiting value and it can be represented by a product of functions
at that location, then the Indeterminate Limit Rule allows you to find those functions one
at a time.

* The smallest dominant term in a product of functions has the property that its natural logarithm shares the
same behavior as the natural logarithm of the entire product at the function’s unbounded location in
question. It also has the property that there is no function within the dominant product other than one that
always outputs 1 that divides into any other term in the product an infinite number of times regardless of
the positive finite power it is raised to, as the unbounded location is approached. These lesser functions can
be apparent or they can be concealed such as they are in   x! as x becomes arbitrarily large.
37

VII. Aux. Theorem I
Aux Theorem I:

If lim f ( x)   , lim g ( x)  0 , and lim g ( x)   (a can be a finite number or it can
xa             x a                xa

represent the end behavior of the function, a =   ), then lim ( f ( x) g ( x))   .
x a

Proof:
Let L = lim g ( x) . If a =  , then for any   0 and R where R is any Real number,
x a

N  0 : x  N , f ( x)  R and L    g ( x)  L   . If a = –  , then for any   0 and
R where R is any Real number, N  0 : x  N , f ( x)  R and L    g ( x)  L   .
Otherwise, if a is finite, then for any   0 and R where R is any Real number, there
exists a delta such that for all x where a    x  a   , f ( x)  R and
L    g ( x)  L   . These conclusions follow directly from the formal definition of
limits. To find an N or a  that simultaneously satisfies f ( x)  R and
L    g ( x)  L   , select the smaller  or the N with the greater absolute value that
satisfies both conditions, if a is finite or infinite, respectively. For all assignments to a,
there is some domain for which f ( x)  R and L    g ( x)  L   . Let  be some
number less than L. Since L is greater than 0, there are an infinite number of Real
numbers between 0 and L. Therefore this assignment to  is possible. If the former
inequality is multiplied through by g(x), f ( x) g ( x)  Rg( x) is obtained. (The inequality
sign does not change because g(x) is always positive over the restricted domain.) Since R
R2
can be assigned any Real number, let R =         where R 2 is a variable that can be
g ( x)
R2
assigned any Real number. The inequality then becomes f ( x) g ( x)               g ( x) or
g ( x)
simply f ( x) g ( x)  R2 by cancellation. Since R 2 can be assigned any Real number, and
the domain for which this inequality is true has not changed, lim ( f ( x) g ( x))   by the
x a

formal definition of limits.
 Aux. Theorem I holds true.

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