# A general limit for exponentials by waabu

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```									                          A General Limit for Exponential Functions

╬
Francis J. O’Brien, Jr., Ph.D.
Aquidneck Indian Council
Newport, RI
March 14, 2013

Calculus is built on the notion of limits
(Finney & Thomas, p. 77)

Introduction

In this paper we demonstrate a general limit for exponential functions by elementary
differential calculus methods. The suggested limit generates a number of useful limits describing
growth or decay1. Numerous examples are provided.
For example, two well-known exponential limits that can be derived from a general limit
are:

n                                  n
 1                                     x
lim 1    e                          lim 1    e x
n     n                             n     n

———
or
———

1                                          1
n                                          n
lim 1  n        e                     lim 1  nx         ex
n 0                                      n0

Note that it is possible to define an infinite limit by a corresponding 0 limit. This is explained
below. The limits in this paper are positive or going in the positive direction, n  , n  0 .
The limit and derivative for a general exponential function a x is also provided.

1
For a good article that describes some practical uses of exponential growth and decay functions, see the website,
http://www.willnichols.me.uk/a/exp/ar01s01s01.html.
1
Standard calculus texts by Bers (pp. 379 ff.) and Finney and Thomas (pp. 481 ff.) as well
as others provide excellent coverage for limits of exponential functions that tend to 0 or  .
limits. Later we indicate a website that calculates limits which is useful to check answers
derived by hand.

——————

Demonstrating the general limit presented next can be done in two ways—(a) by log
differentiation (with a small “trick”) and (b) the more general method of L’Hôpital’s Rule2 for
differentiating “indeterminant forms” of fractions of continuous functions. We recommend the
YouTube lecture MIT 18.01 (Lecture 35, Prof. David Jerison) for an excellent detailed
explanation of the mathematical approach presented to undergraduate students of calculus.
Search for “MIT 18.01 Lec 35”.

General Exponential Limit

A general limit for simple exponential functions is:

bn
 a
lim 1                     e ab ,
n     n
where a & b are some arbitrary fixed constants.

NOTE: As indicated below, the limit can be expressed in alternative ways by re-arranging the
a, b parameters. We use the one stated above.

Log Differentiation

The first approach to derive the general limit e ab is log differentiation with a clever trick.
Let the limit be called y. Take the natural log of y, re–define the limit to go to 0, substitute,
differentiate, and get y back by exponentiation. Applying the steps,

2
An alternate spelling for this French name is L’Hospital.
2
bn
 a
(1) y  lim 1  
n     n
 a
(2) ln y  b lim n ln1  
n        n
a                   a
(3) Let  x  0, n           , and substitute
n                  x
ln 1  x 
(4) ln y  ab lim
x  0      x

From the limit in step (4) the “trick” is to add the quantity ln 1 = 0 in the numerator which does
not change the limit,

ln1  x   ln 1
ln y  ab lim                       .
x  0        x

This difference quotient now looks like a log derivative with a constant outside. If we recall the
definition of the derivative of ln x,

d               ln x  x   ln x 1
ln x  lim                       ,
dx       x  0         x          x

we see that,

ln 1  x   ln1
lim
x  0         x

is the derivative of ln x evaluated at x  1 ; i.e.,

ln 1  x   ln 1 d
lim                        ln x       1.
x 0           x          dx    x 1

Substituting in (4),

ln 1  x   ln1
ln y  ab lim                        ab.
x  0         x

3
Thus, ln y  ab.         Lastly, by definition, y  e ln y  e ab , which proves the limit,
bn
 a               ab
lim 1          e        .
n     n

NOTE: The reader can verify that the general limit just derived can be expressed in other ways
such as:

n                 abn
 ab         1                       ab
lim 1    lim 1                  e        .
n     n  n     n

Apply log differentiation to demonstrate.

L’Hôpital’s Rule

This x  0 log differentiation method is straightforward but does not work for all limits.
1

When it does not work—such as for lim 1  n  —the method known as L’Hôpital’s Rule is
n
n
the logical alternative to try. L’Hôpital’s Rule is applied by differentiating “indeterminant”

f ( x)
forms of fractions of two continuous functions                . For example, the limit
g ( x)

f ( x)       1 ex
lim           lim
x  0 g ( x) x  0 x

1  e0        0
reduces to lim         lim     if the limit 0 is substituted. This meaningless limit is called
x 0 0        x 0 0
“indeterminant” and cannot be evaluated in this form.
The functions f ( x) & g ( x) of an indeterminant form are differentiated independently in
the L’Hôpital method of limit evaluation,

4
d
f ( x)
f ( x)                         f ( x)
lim          lim dx         lim            .            L’Hôpital’s Rule
x  a g ( x) x  a d           x  a g ( x )
g ( x)
dx

The above limit by the method of L’Hôpital’s Rule is:

d
(1  e x )
f ( x)       1 ex
lim          lim        lim dx              lim e x   lim 1  1.
x 0 g ( x) x  0 x       x 0    d            x 0        x 0
x
dx

The following plot shows the limit clearly (Fig. A) by inspection of the curve at x  0.

x

1 ex
Fig. A.          approaches –1.
x
Source: Wolfram Mathematica, http://www.wolframalpha.com/

NOTE: As textbooks point out, L’Hôpital’s Rule may be applied repeatedly until a clear answer
appears provided that after each round of differentiation you still have an indeterminant form
f ( x)
based on the original functions               .   Sometimes L’Hôpital’s Rule cannot find the limit in
g ( x)
which case other means are required such as graphical plots, series expansions or other methods.

Common indeterminant forms of functions f ( x) & g ( x ) that can be analyzed by
L’Hôpital’s Rule (whether they be combined products, fractions, sums or differences for
algebraic or transcendental functions) include the types of:

5
0 1 
, , ,   , 0 0 ,  0 ,   , 0  , 1 . (See Bers, pp. 132 ff. and pp. 547 ff. for rules,
0 0 
guidelines and examples in dealing with infinite limits3). Also, the “type” can change if the
original functions in the form of a product f ( x)  g ( x) for instance need to be re-expressed to
f ( x)
create a fraction of an indeterminant form                       . For example, the limit type
g ( x)
ln( x)                                         
0   lim e  x ln( x) when expressed as lim               becomes indeterminant of type           .
x                                    x  e x                                           
L’Hôpital’s Rule, by differentiation and the Product Rule for limits, gives the answer of 0
ln( x)
indicating that e x grows faster than ln( x) as x increases. A plot of           by x will show this
ex
intuitively (Fig B., below).

x

ln( x)
Fig. B.            approaches 0.
ex
Source: Wolfram Mathematica, http://www.wolframalpha.com/

NOTE: L’Hôpital’s Rule is a delicate tool which must be used with great care. It is important to
understand when L’Hôpital’s Rule is not appropriate (or necessary). For example, consider the
limit,

1  2x
lim        .
x 0 2  4x

1
Setting the limit x to 0 in the fraction gives the answer    without further analysis. This means
2
the limit of the fraction is not “indeterminant” and, therefore, L’Hôpital’s Rule is not valid. If
L’Hôpital’s Rule is applied mistakenly to the limit by differentiating numerator and denominator

3
But some infinite limit forms of functions f ( x ) & g ( x ) are not indeterminant such as :
    , ()()  , ()()  . As one example,
lim e x ln( x)  lim e x lim ln( x)      .
x                   x       x 
6
1
terms separately, the incorrect answer of          results (example from Bers, p. 548).
2

Applying L’Hôpital’s Rule

bn
 a
The general limit for y  lim 1   is turned into a fraction of indeterminant form by
n     n
taking the log of the limit and adjusting as needed to form a fraction of functions. This is a
standard approach for a limit that has an exponent outside. For this method, we start above at
step (2),

 a
ln y  b lim n ln1   ,
n       n

and, since we need a fraction of an indeterminant form to apply the L’Hôpital method,
algebraically flip the n outside the log term to form the equivalent expression,

 a
ln1  
ln y  b lim 
n
.
n      1
n

This common operation for exponential limits gives us an indeterminant fractional form of type

          a               
     ln1  
  ln 1        0
i.e., 
0
            .
0          1      0          0

                           


Now, take separate derivatives with respect to n of the numerator and denominator functions
d  a
f (n) and g (n) ; the chain rule is needed for numerator term,    ln1   . The result by
dn  n 
L’Hôpital’s Rule will be,

7
d 1
a
dn n
d  a                a
ln1           1
f ( n)         dn  n               n  ab lim 1  ab,
lim          b lim             b lim
n   g ( n)    n     d 1        n  d 1      n    a
1
dn n            dn n            n
or, ln y  ab, and y  e ab .

NOTE: The alternate forms of the e ab limit mentioned earlier can be derived by L’Hôpital’s
Rule.
ab
Thus, both log differentiation and L’Hôpital’s Rule can prove the general limit e .
When it can be applied, log differentiation is usually the easier method to employ, especially for
complicated limits (exemplified below in the last section).

L’Hôpital’s Rule, however, is the more general approach—for example, for relative
xn
growth comparison of two functions such as, lim                  case. This limit is derived
, which is a
x  e x           
below in the Appendix using a number of clever manipulations. Some limits of indeterminant
form can be done without either method. See O’Brien (2013) who analyzes a complicated limit
xn
lim     using only Laws of Exponents and limit rules.
n   n!

bn
 a        ab
NOTE: to obtain the equivalent limiting form of lim 1    e for the limit going to 0,
n    n
1                                                  1
we re-define by letting h  , which indicates the new limit will now be h  lim  0, so
n                                             n  n
that, re- writing the limit,

1                      b
bn   h  0
      a           n                                ab
lim 1  ah 
h
lim 1                                           e        .
n        n                  h 0

Applying the 4 steps of log differentiation above will show the result. Now, if desired we can
b
n     ab
re–state the result in terms of n instead of h, lim 1  an   e . L’Hôpital’s Rule for the
n 0
0
log of the limit shows the same result for indeterminant form . This process was done to
0
8
obtain the 0–limits of e & e x at the beginning of the paper.
NOTE: To obtain the decay form e  ab we let a  a in the general limit e ab and proceed as
follows:

bn
 a
(1) y  lim 1  
n   n 
 a
(2) ln y  b lim n ln1  
n        n
a                      a
(3) Let   x  0, n  
n                     x
ln 1  x   ln 1
(4) ln y   ab lim                          ab
x  0         x

This leads to y  e ln y  e  ab .        L’Hôpital’s Rule also shows the answer by creating the
indeterminant fractional form from the limit as explained above for e ab .
1
By the same reasoning shown earlier, for the same limit going to 0, set h                             0,
n

1                       b
bn     h  0
 a               n                                 ab
lim 1  ah 
h
lim 1                                          e          ,
n     n                    h0
or in terms of n ,

b

 e  ab ,
n
lim 1  an 
n 0

by log differentiation or L’Hôpital’s Rule.

Other Limits

We can obtain a number of other limits for elementary exponential functions from the
general limit,

b
bn
 a
lim 1  an 
n
lim 1           e ab            or                                     e ab
n     n                                   n 0

9
by substitution of parameters in e ab and adjusting the terms in the limit expression to satisfy the
desired result. Sometimes this involves trial and error.

NOTE: An intuitive “trick” to verify and create these elementary limits for exponential
bn
 a
functions is to recognize that in, lim 1                     e ab , we can look at the product of the inside
n     n
 a
term and the exponent, and by algebra,   bn   ab (the n terms cancel) which is
 n
the ln y term in the derivation . The trick works when the base term of the function is of the form
1  and the n terms cancel. It would not work on the limit lim 1  n n which never converges,
a
n                                                        n 

as a simple plot of the function 1  n n will show4.
b

The “trick” is useful also for the 0-limit form, lim 1  an 
n
 e ab by the same principle,
n0
n
 an  b   ab .
 

But it would not work for lim 1 
1
  1.
n                                     n  0             n

Examples by General Limit

The following examples are not derived but they can be checked by hand calculations5.

bn
 a
   If we want e  ab , set a   a in lim 1                     e ab or
n     n
b
bn
 a
 lim 1  an         e  ab . This was shown earlier.
n
lim 1  
n     n             n 0

o NOTE: switching the roles of a & b will give the same limit of e  ab as the reader
can verify

4
The Sandwich Theorem for Sequences (Finney and Thomas, pp. 580 ff.) is an advanced method which will
evaluate limits of this kind. Or, the log of the limit shows lim n lim ln1 n   .
n  n 
5
One excellent online limit calculator is found at the website, http://www.numberempire.com (“Limit Calculator”).
10
b                                                bn b
1               1 
   The limit e a is obtained by making a  , since lim 1    e a .
a       n     an 

o NOTE: The reader can verify that the following limit going to 0 is also
b         b
          n n
correct: lim 1            ea .
n  0          a

        The decay forms of the limit are:

b    b
   1 
bn
  n n  
lim 1                 lim 1    e a .
n     an               n  0  a

bn
2                                      b                             2
   The limit eb is obtained by setting a  b or lim 1                          eb . For the decay form,
n    n
b
bn
 b                     2
 e  b . The limit going to 0 is lim 1  nb          e b .
n         2
use lim 1  
n     n                                                  n 0

n
 1
   The well-known limit, lim 1    e  2.718281828459045... results by setting
n     n
bn
 a
a  b  1 in lim 1    e ab . This is a definition of e that can be used to compute e
n     n
to as many decimal places as desired for large n. Another formula comes from the
1 1 1            1
(infinite series) expansion, e  2       for n as large as desired.
2! 3! 4!         n!
1
n
 x
lim 1    lim 1  nx   e x is achieved by setting a  x, b  1 in
n

n     n   n 0
bn
 a
lim 1             e ab .
n     n
1
n
 x
o The decay form is lim 1    lim 1  nx   e  x .
n
n    n   n 0

11
1
n
                         x
  lim 1  nx                  e  x can be
n
NOTE: The reader can verify that lim 1 
n                           n   n 0
written equivalently as,

x
nx
 1
 lim 1  n             e x .
n
lim 1  
n     n           n 0

NOTE: The limiting values of e  x in limit calculations are found by simple inspection;
lim e x  , lim e  x  0, lim e  x  1, etc.
x                  x                x 0

anx k 1
A more complicated limit lim 1  
x                                          k
                                                                         e ax       for any constants a & k is done by
n     n
trial and error.                   In the general limit we initially set                            a  x, b  a which gives
an
 x
e ax  lim 1                   at this point.    Next we multiple both sides by                             x k 1 to give
n     n
x k 1            k
 e ax 
                   e ax        which by the Laws of Exponents is the desired limit. By the intuitive
      
 x
 n
k
“trick” it seems correct since the algebraic product    anx k 1  ax k  y  e ax .        
anx k 1
o The decay form of the limit is lim 1  
k
 e  ax . Here the “trick”
x
     
n   n 
 x
       
indicates    anx k 1   ax k  y  e  ax . Analysis will show it to be
 n
k

correct.

1
nx
2         1
 x                       x2
      For example, lim 1                     e 2             results from setting
n    n

k
a  , k  2 in e  ax .
1
2

12
k
o The 0–limit forms are derived by flipping n in e  ax to obtain the positive and
negative exponential forms:

ax k 1                                 ax k
k
lim 1  nx                           lim 1  n 
n                                     n
or                                e ax
n 0                                   n 0
ax k 1                                 ax k
k
lim 1  nx                            lim 1  n                 e  ax
n                                      n
or
n 0                                    n 0

a  r nx k 1              a  r  x k
  x                                        
   In the same manner as above, lim 1  t                                   e            t           .
n      n 

      
o For the 0-limits, flip the n terms.

   The usefulness of the 4-step log differentiation method is appreciated by deriving the
limit,

bn     ab
      a 
lim 1                 e k .
n      nk  t 
a            1 a    
o Set x              0, n      t .
nk  t         k  x 

Then, derive the limit using L’Hôpital’s Rule which requires more than one application
of the Rule. The 0-limit (by flipping the n terms) using L’Hôpital’s Rule will also be
challenging in comparison to log differentiation.

   A general exponential function a x a  0 has a limit similar to e x which is a special case
of a x . We can use the limit for e x to get the a x limit. By definition, a  e ln a , and by
Laws of Exponents,

 
a x  e x ln a  e x
ln a
.

13
Recall,

n
 x
e x  lim 1   .
n   n 

Then,

ln a
ln a                n                    n ln a
ex                 x                     x
         lim 1              lim 1              , or
          n    n              n   n 
           
n ln a
 x
a x  lim 1              .
n   n 
Note that if a is set to e the limit for e x results from a x since ln e  1 (which can be
shown by log differentiation or L’Hôpital’s Rule for the limit of e).

o The 0–limit form is obtained by flipping n in the infinite limit,

ln a                            x ln a

a x  lim 1  nx                      lim 1  n 
n                                 n
or                              ,
n0                               n0

and, the decay form is,

          x
n ln a
 lim 1  
1 n   n 

a x                   ln a
ax 
 lim 1  nx 
n
n  0


Log differentiation or L’Hôpital’s Rule will show the limits for a x and a  x as correct.

NOTE: The reader can verify by L’Hôpital the following related limits:

14
 ah  bh 
     lim            ln a  ln b  ln a  .
 

h  0    h                       b


 ah 1
      lim         ln a
 h 
h  0      

o From this limit we can obtain the derivative:

d x        a xh  a x
a  lim
dx    h0       h
ah 1
 a x lim          a x ln a,
h0       h
d
and, if a  e, a x  e x ln e  e x .
dx

   Continuous compound Interest. The reader is invited to look up the formula for
computing continuous compound interest and inspecting the infinite limiting form; try to
determine the exponential form it takes by substituting parameters for the general limit
derived in this paper.

15
APPENDIX
xn
lim         0
x  e x

Proof of Limit by L’Hôpital’s Rule

In this appendix we show a proof of this interesting limit by the method of L’Hôpital’s

Rule for indeterminant forms by derivatives. The limit is obviously indeterminant of type

and cannot be evaluated directly. The limit expression shows that x is a moving variable and n is
any fixed constant no matter how small or large, say a positive integer, n  1,2,3,....

An intuitive approach to observing the limit behavior begins by plotting the ratio of the
x2
functions. Figure 1 is a plot showing the limit for n  2 . It is clear that as x increases, lim
x  e x
goes to 0. Larger values of n can be tried with the same result.

x2
y
ex

x

x2
Figure1. Plot of            which tends to 0.
ex
Source: Wolfram Mathematica, http://www.wolframalpha.com/.

Two approaches can be made to show the limit by L’Hôpital’s Rule for n  2 or any n—
a hard way and an easy way. For n  2 one way is by taking successive derivatives of the
functions x 2 & e x until an indeterminant form no longer exists and the limit can be clearly
evaluated. The first application of L’Hôpital’s Rule shows,

16
x2
d 2
x      2x
lim    = lim dx    lim      .
x  e x x  d x    x  e x
e
dx
                                          2x
Since the fraction is still      , we use L’Hôpital again on the limit of    ,
                                          ex

2x
d
2 x           1  2
lim     lim dx         2 lim      0.
x  e x x   d e x       x  e x 
dx

x2
This demonstration shows that lim                 0 after two applications of L’Hôpital’s Rule .
x  e x
x100
But what if we had to evaluate lim                     ? That proof would require 100 individual
x    ex
derivatives of x100 before we obtained a final limiting value which cannot be predicted
beforehand. Surely an easier approach exists.
The second approach is sophisticated and uses multiple rules for limits, Laws of
Exponents, and derivatives. We state each step as we go along to show the development of the
derivation.
To show the easier method6, we first restate the limit in an equivalent form,

n
xn         x 
lim      lim        

x  e x   x   e x n 

which is correct by the Laws of exponents. Since n is any integer we can expand the limit n
times,

n 
times
         
n          x  x   x 
xn        x 
lim     lim                    lim                 
 x n  x n  x n  .
x  e x x   e x n 
                   x   e
       e    e    

x
6
Another approach is by inverse functions. Set y  e  , and x  ln y , and by L’Hôpital’s Rule solve the
n
 ln y 
limit, lim          , as shown in text by integer expansion .
y  y1 / n 
        

17
By the Product Rule for limits, this can be expressed as a product of n limit terms,

n
xn        x 
lim     lim 



x   e x x   e x n 

n 
 times
             
 x  x   x 
 lim                       
x    e x n  e x n   e x n 

                        
n 
times               
 x                                 
 lim         lim  x   lim  x 
x   e x n  x    e x n  x    e x n 
                                   

Note that this last expression is n indeterminant forms versus the one we started with. It is
legitimate now to apply L’Hôpital’s Rule to each individual limit; that is, differentiate the

 d                  
    x          1              
functions separately, lim  dx       lim         lim  n  using the Chain Rule
 xn
x   d x n  x   1 x n  x    e
 e             e               
 dx            n    
x
d
for e n . Do this for each of the n product terms. Then, collect the constant terms, and restate
dx
all of the n limit terms as one limit term by the Laws of Exponents for final evaluation.
Repeating all steps:

18
n
xn        x 
lim     lim       
xe  x x   e x n 
      
n 
 times
             
 x  x   x 
 lim                       
x    e x n  e x n   e x n 
                      
          n times
               
 x                                        
 lim           lim  x          lim  x              Apply L' Hôpital here
x   e x n

 x   x n
      e
 x   x n
       e


n
 times 
                       
                                       
                                       
1               1               1 
 lim              lim            lim 
x   1 e x n  x   1 e x n  x   1 e x n 
                                       
n               n              n       
n
 n                       n
 lim          n n lim  1   n  0
                   
x   e x n       x   e x  

xn
This concludes the proof we sought,                    lim            0,   which demonstrates the power of
x  e x
L’Hôpital’s Rule for finding an answer in a much faster way.

_________

p
NOTE: If n is a fraction of form          , q  0, then the following limit can be solved per above
q
procedure,
p q
 xp q                         
lim          lim                  
x
x    x  x    q p  x           
 e           e                

The reader can work out the details to show:

p q
 xp q                        
lim         lim                            p q  p q lim
x                                           1
 0.
x   x  x   q p x                                   x  e x
 e           e               

p
Hint: expand the limit algebraically          times.
q
19
Other approaches are possible including the inverse relations method, y  e x  .

References

Bers, L. Calculus, Vol. 1, 1969. Holt, Rinehart and Winston, Inc.

Finney, R.L. and G.B. Thomas, Jr. Calculus, 1990. Addison-Wesley Publishing Co.

“Limit Calculator”. http://www.numberempire.com.
MIT 18.01. Single Variable Calculus, Fall 2007, Lecture 35: Indeterminate forms— L'Hospital's

xn
O’Brien, F. J. Jr. Proof that lim       0 by Stirling’s Approximation for Factorials, 2013.
n   n!
http://www.docstoc.com/profile/waabu.

xn
__________. Proof that lim       0 by Stirling’s Approximation for Factorials (Part II), 2013.
n   n!
http://www.docstoc.com/profile/waabu.