# Limits

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```					                    SQQM1023: Managerial maths

Limits

5.1 Introduction

Perhaps you have been in a parking-lot
situation in which you must “inch up” to the
car in front, but yet you do not want to bump
or touch it. This notation of getting closer and
closer to something, but yet not touching it, is
very important in mathematics and is involved
in the concept of limits.

Basically, we will let a variable “inch up” to a
particular value and examine the effect it has
on the values of a function.

Definition

The limit of f(x) as x approaches c is the
number L, written
lim f ( x)  L
x c

provided that f(x) is arbitrarily close to L for
all x sufficiently close to, but not equal to, c.

We emphasize that, when finding a limit, we
are concerned not with what happens to f(x)
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when x equals c, but only with what happens
to f(x) when x is close to c.

Moreover, a limit must be independent of the
way in which x approaches c. That is, the limit
must be the same whether x approaches c
from the left or from the right(for x < c or
x > c, respectively).

5.2 Estimating a Limit from a Table

For example, consider the function
x3  1
f ( x) 
x 1

Although this function is not defined at x = 1.
When x takes on values closer and closer to 1,
regardless of whether x approaches () it
from the left(x < 1) for example x = 0.8, 0.9,
0.95, 0.99,… or from the right (x > 1) for
example 1.2, 1.1, 1.05, 1.01, … , the
corresponding values of f(x) get closer and
closer to one and only one number, 3. This is
also clear from the table below:

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x<1                                       x>1
x             f(x)                          x           f(x)

0.8      2.44                1.2       3.64
0.9      2.71                1.1       3.31
0.95     2.8525              1.05     3.1525
0.99     2.9701              1.01     3.0301
0.995     2.9850             1.005     3.0150
0.999     2.9970             1.001     3.0030
Therefore, we say that the limit of f(x) as x
approaches 1 is 3 and write
x3  1
lim        3
x 1 x  1

Example 3:

Use your calculator to complete the table, and
use your results to estimate the given limit.
1
x  3  x  3
x           2.9    2.99     2.999 2.9999
f x

x        3.1             3.01            3.001     3.0001
f x

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x2  4
b) lim
x 2 x  2
x   -2.1 -2.01           -2.001 -1.999          -1.99       -1.9
f x

5.3 Estimating a Limit from a Graph

For example, estimate lim f ( x) , where the
x1
graph of f is given in figure below:
y

5

3

1

0       1                   3                x

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Solution:

If we look at the graph for values of x near 1,
we see that f(x) is near 3. Moreover, as x gets
closer and closer to 1, f(x) appears to get
closer and closer to 3. Thus, we estimate that
lim f ( x) is 3
x1

Example 4:

Refer figure above, estimate lim f ( x)
x0

Exercise 1.1

Estimate lim f ( x) if it exists, where the
x  2
graph of f is given in figure below

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y

y = f(x)

3

2

1

x
-2

5.4 Using Properties to Determine Limits.

Properties of Limits

1. If f(x) = c is a constant function, then
lim f ( x)  lim c  c
x a              x a

2.   lim x n  a n , for any positive integer n.
x a

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Example,

a)    lim 7  7
x 2

b)    lim k  k where k is constant.
x 5

c)    lim x 2  32  9
x 3

d)    lim x5  25  32
x 2

If L , M , k, are real numbers and
lim f ( x)  L and lim g ( x)  M , then
x a                     x a

3.    lim ( f ( x)  g ( x))  L  M
x a

4.    lim ( f ( x)  g ( x))  L  M
x a

5.    lim ( f ( x).g ( x))  L.M
x a

6. lim (kf ( x))  kL
x a

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Example,
       
a) lim x 2  x  42  4
x 4

= 16  4
= 20

            
b) lim x3  x  4  13  1  4
x 1

= 11 4
= 2

c)   lim  x  2 x  5  3  2  3  5
x 3

= 1  8
=8

d)
x 2

lim  x  12 2 x  4  ?

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e)   lim 2 x  24  8
x 4

f)   lim 3 x 2  ?
x 5

f ( x) L
5. Quotient rule                lim         ,M  0
x a g ( x)  M

6. Power rule
r        r
lim    ( f ( x)) s      Ls , s   0
x a

Example,

x  2 3  2  1     1
a)   lim                    
x 3 x  8 3  8  5    5

x2  5
b) lim         ?
x 2 1  x

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x 2  x  2  3 2 2  2  21
3
c)   lim
x 2

= 3 4  2  21
= 3 27  3

d)   lim        9  x2  ?
x2

5.4.1 Limits and Algebraic Manipulation

We now consider limits to which our limit
properties do not apply and which cannot be
evaluated by direct substitution. Our technique
will be to algebraically manipulate f(x) so as to
obtain a form to which our limit properties will
apply.

Example, find

3x 2  7 x  2
a) lim                 ?
x 2     x2

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x4
b)   lim 2        ?
x  4 x  16

5.5 One-Sided Limits

Consider the graph of a function f,

y

1

x
-1

Notice that f(x) is not defined when x = 0. As
x approaches 0 from the right, f(x) approaches
1. We write this as
lim f ( x)  1
x 0 

On the other hand, as x approaches 0 from the
left, f(x) approaches –1 , and we write

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lim f ( x)  1
x 0 

Limits like these are called one-side limits.

5.6 Limits at Infinity

Now let us examine the function
1
f ( x) 
x

as x becomes infinite, first in a positive sense
and then in a negative sense. From table
below:

x        f(x)                          x      f(x)
1000    0.001                      -1000    -0.001
10,000 0.0001                      -10,000  -0.0001
100,000 0.00001                    -100,000 -0.00001
1000000 0.000001                   -1000000 -0.000001

x increases without bound through positive
values, the values of f(x) approach 0.
Likewise, as x decreases without bound

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through negative values, the values of f(x) also
approach 0.

Symbolically, we write
1                1
lim  0 and lim        0
x  x         x   x

In general,
1               1
lim p  0 and lim p  0 where p > 0
x  x        x   x

Example 5:

Find the limit (if it exist).
4x2  5
lim
x  2 x 2  1

Solution:
To do this, we divide both the numerator and
denominator by the greatest power of x that
occurs in the denominator. Here it is x2. This
gives:

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4x2
5
 2
4x2  5         x2 x
lim          lim
2
x  2 x  1 x  2 x 2    1
 2
x2 x
5                      5
4  2 lim 4  lim 2
 lim      x  x            x  x
x      1                      1
2  2 lim 2  lim 2
x    x           x  x
40
          Since 1
20             lim
x  x p
0

2

5.7 Limits for a Compound Function

Example 6:

x2  1            x 1
If f ( x)                                   , find the limit
3                 x 1
(if it exist)

a)    lim f ( x)                        b)         lim f ( x)

x 1                                          x 1

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c)   lim f ( x)
x1

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