LIMITS
OF
FUNCTIONS
LIMITS OF FUNCTIONS
OBJECTIVES:
• define limits;
• illustrate limits and its theorems; and
• evaluate limits applying the given
theorems.
• define one-sided limits
• illustrate one-sided limits
• investigate the limit if it exist or not using
the concept of one-sided limits.
• define limits at infinity;
• illustrate the limits at infinity; and
• determine the horizontal asymptote.
DEFINITION: LIMITS
The most basic use of limits is to describe how a
function behaves as the independent variable
approaches a given value. For example let us
2
examine the behavior of the function f ( x )  x  x  1
for x-values closer and closer to 2. It is evident from
the graph and the table in the next slide that the
values of f(x) get closer and closer to 3 as the values
of x are selected closer and closer to 2 on either the
left or right side of 2. We describe this by saying
that the “limit of f ( x )  x 2  x  1 is 3 as x
approaches 2 from either side,” we write
lim  x  x  1   3
2
x 2
y
y  x  x1
2
f(x)
3
f(x)
O
x
2
x
1.9
1.95
1.99
1.995 1.999
F(x)
2.71
2.852
2.97
2.985 2.997
left side
2
2.001 2.005
2.01
2.05
2.1
3.003 3.015 3.031 3.152
3.31
right side
This leads us to the following general idea.
1.1.1 (p. 70) Limits (An Informal View)
EXAMPLE
Use numerical evidence to make a conjecture about
x1
the value of lim
.
x1
x 1
x1
Although the function f ( x ) 
is undefined at
x 1
x=1, this has no bearing on the limit.
The table shows sample x-values approaching 1 from
the left side and from the right side. In both cases the
corresponding values of f(x) appear to get closer and
closer to 2, and hence we conjecture that
x1
lim
 2 and is consistent with the graph of f.
x1
x 1
Figure 1.1.9 (p. 71)
x
.99
.999
.9999
.99999
F(x) 1.9949 1.9995 1.99995 1.999995
1
1.00001
1.0001
1.001
2.000005
2.00005 2.0005
1.01
2.004915
THEOREMS ON LIMITS
Our strategy for finding limits algebraically has two
parts:
• First we will obtain the limits of some simpler
function
• Then we will develop a list of theorems that will
enable us to use the limits of simple functions as
building blocks for finding limits of more
complicated functions.
We start with the following basic theorems,
which are illustrated in Fig 1.2.1
1 . 2 . 1 Theorem
a 
Let a and k be real numbers.
lim k  k
x a
Theorem 1.2.1 (p. 80)
b 
lim x  a
x a
Figure 1.2.1 (p. 80)
Example 1.
If f  x   k is a constant
remain
function, then the values of f(x)
fixed at k as x varies, which explains why
f(x)  k as x  a for all values of a.
For example,
lim 3  3
x  -25
lim 3  3
x0
lim 3  3
x
Example 2.
If f  x   x, then x  a it must also be true that f  x   a .
For example,
lim x  0
x0
lim x   2
x  -2
lim x  
x
The following theorem will be our basic tool for
finding limits algebraically.
Theorem 1.2.2 (p. 81)
This theorem can be stated informally as follows:
a)
b)
c)
d)
The limit of a sum is the sum of the limits.
The limit of a difference is the difference of the limits.
The limits of a product is the product of the limits.
The limits of a quotient is the quotient of the limits,
provided the limit of the denominator is not zero.
e) The limit of the nth root is the nth root of the limit.
• A constant factor can be moved through a limit symbol.
EXAMPLE : Evaluate the following limits.
1 . lim  2 x  5   lim 2 x  lim 5
x 4
x 4
x 4
2 . lim 6 x  12   lim 6 x  lim 12 
x 3
x 3
 6 ( 3 )  12
 2 lim x  lim 5
x 4
x 4
 18 - 12
 2( 4 )  5
6
85
 13
3 . lim 4  x ( 5 x  2 )  lim 4  x   lim 5 x  2 
x 3
x 3
x 3



 lim 4  lim x   5 lim x  lim 2 
 lim 4  lim x  lim 5 x  lim 2
x 3
x 3
x 3
x 3
x 3
x 3
 4  3 5 ( 3 )  2 
 1 13 
 13
x 3
x 3
x 3
4 . lim
2x
x 5


6 . lim
5x  4
lim 2 x
lim  5 x   lim  4 
2 5 
25  4
x 5


2 lim x
x 5
5 lim  x   lim  4 
x 5
x 5
21

5 . lim 3 x  6   lim 3 x  6 
lim
x 1
x3

9
3
x 3
4

3
lim 3 x  lim 6 
 3 lim x  lim 6 

3
x 3
x 3
3
x 3

x 3
 3  3   6 
 3375
3
 15 
3
8x 1

10
3
x 3
x3
x1
x 5
x 5
8x  1

2
OR
When evaluating the limit of a function at a
given value, simply replace the variable by
the indicated limit then solve for the value of
the function:


lim 3 x  4 x  1  3  3   4  3   1
x3
2
2
 27  12  1
 38
EXAMPLE: Evaluate the following limits.
x 8
3
1 . lim
x  2
x2
Solution:
x 8
3
lim
x  2
x2

 2 3
8
22
88


0
0
(indeterminate)
0
Equivalent function:
 lim
 x  2 x 2  2 x  4 
x2
2
 lim x  2 x  4
x  2
x  2


  2   2  2   4
2
 4  4  4  12
x 8
3
 lim
x  2
x2
 12
Note: In evaluating a limit of a quotient which
reduces to 0 , simplify the fraction. Just remove
0
the common factor in the numerator and
denominator which makes the quotient 0 .
0
To do this use factoring or rationalizing the
numerator or denominator, wherever the radical is.
x2 
2 . lim
x0
2
x
Solution:
x2 
lim
x0
2
02

x
2

0
0
(indeterminate)
0
Rationalizing the numerator:
x2 
 lim
x0
 lim
x0

x
 lim
x0
2
x

x
x2 
2
x2 
2
x

x2 
2
x2 
2
 lim
x0

2
4
 lim
x0
1
x2 
x


2
x22
x2 
2
1
2

2

1
2 2

2
4
8 x  27
3
3 . lim
x
4x  9
2
3
2
Solution:
3
8 x  27
3
8    27
2
3
lim3
x
4x  9
2
2

27  27

2
3
4   9
2
99

0
(indeterminate)
0
By Factoring:
 2 x  3 4 x 2  6 x  9 
 2 x  3  2 x  3 
 lim
x
3
2

2
999

33
27

6
8 x  27
x
3
2
4x  9
2

2
3
 lim
9

3 2
2
4x  6x  9
2
 lim
x
2 x  3 
3
2
3
2

3 2
2

3
3
4   6    9
2
2
 3

 2   3 
 2

x  2x  3
3
4 . lim
x 5
2
x 2
Solution:
lim
x  2x  3
x2
x 5
2
2  2 2  3
3
3



2  5
2
843
45
15
9

15
3
x  2x  3
3
 lim
x 2
x 5
2

15
3
DEFINITION: One-Sided Limits
The limit of a function is called two-sided limit if
it requires the values of f(x) to get closer and closer
to a number as the values of x are taken from
either side of x=a. However some functions exhibit
different behaviors on the two sides of an x-value a
in which case it is necessary to distinguish whether
the values of x near a are on the left side or on the
right side of a for purposes of investigating limiting
behavior.
Consider the function
1
x 0
x0
As x approaches 0 from the right, the
values of f(x) approach a limit of 1, and
similarly , as x approaches 0 from the
left, the values of f(x) approach a
limit of -1.


 1,
f(x)
 
x
  1,
x
-1
In symbols ,
lim
x o
x
x
1
and
lim
x o
x
x
 1
This leads to the general idea of a one-sided limit
1.1.2 (p. 72) One-Sided Limits (An Informal View)
1.1.3 (p. 73) The Relationship Between One-Sided and Two-Sided Limits
EXAMPLE:
1. Find if the two sided limits exist given
f(x)
x
x
SOLUTION
x
lim
x o
1


1
x
sin ce the
then
the
and
lim
x o
x
lim
x o
 lim
or lim
x o
x
x
 1
x
x o x
x
two sided lim it does not exist
-1
x
x
does not exist.
EXAMPLE:
2. For the functions in Fig 1.1.13, find the one-sided
limit and the two-sided limits at x=a if they exists.
SOLUTION
The functions in all three figures have the same
one-sided limits as x  a , since the functions are
Identical, except at x=a.
These
lim its are
lim  f ( x )  3
x a
and
lim  f ( x )  1
x a
In all three cases the two-sided limit does not exist as
x  a because the one sided limits are not equal.
Figure 1.1.13 (p. 73)
EXAMPLE:
3. Find if the two-sided limit exists and sketch the graph of
 6 + x if x < -2 
g(x) =  2

x
if
x

-2


SOLUTION
a . lim  g ( x )  lim
x  2
x  2

6
 x
b . lim  g ( x )  lim  x
x  2
x  2
6 2
 - 2 
 4
 4
2
sin ce the lim g ( x )  lim g ( x )
x  2

x  2

then the two sided lim it exist and is equal to 4
or lim g ( x )  4
x  2
2
y
4
-6
-2
4
x
EXAMPLE:
4. Find if the two-sided limit exists and sketch the graph
of
 3 + x 2 if x < -2 


f (x) = 
0 if x = -2 


2
1
1
x
if
x
>
-2


and sketch the graph.
SOLUTION
a . lim
x  2

f ( x )  lim
x  2
3  x 
2

 3   2 
b . lim
x  2

x  2
 11  - 2 
2
7
2
7
sin ce the lim
x  2

f ( x )  lim
then the two sided
or

f ( x )  lim  11  x
lim f ( x )  7
x  2
x  2

f(x)
lim it exist and is equal to 7
2

EXAMPLE:
5 . If f ( x )  3  2 x  4 , det er min e if lim f(x) exist ,
x2
and sketch the graph.
SOLUTION
a . lim  f ( x ) 
x 2
b . lim  f ( x ) 
lim  3  2 x  4
x 2
x 2
lim  3  2 x  4
x 2
 3  2 2   4
 3  2 2   4
 3
 3
sin ce the lim  f ( x )  lim  f ( x )
x 2
x 2
then the two sided lim it exist and is equal to 3
or
lim f ( x )  3
x 2
f(x)
(2 ,3 )
2
x
DEFINITION: LIMITS AT INFINITY
If the values of the variable x increase without
bound, then we write x   , and if the values of
x decrease without bound, then we write x   .
The behavior of a function f ( x ) as x increases or
decreases without bound is sometimes called the
end behavior of the function.
For example ,
lim
x  
1
x
0
and
lim
x  
1
x
0
x
x
lim
x  
1
x
0
lim
x  
1
x
0
In general, we will use the following notation.
1.3.1 (p. 89) Limits at Infinity (An Informal View)
Fig.1.3.2 illustrates the end behavior of the function f when
lim f ( x )  L or lim f ( x )  L
x  
Figure 1.3.2 (p. 89)
x  
EXAMPLE
Fig.1.3.2 illustrates the graph of
Figure 1.3.4 (p. 90)
x
1

y  1   .
x

As suggested by
this graph,
1

lim  1  
x  
x

x
1

lim  1  
x  
x

x
 e and
 e
EXAMPLE ( Examples 7-11 from pages 92-95)
1 . lim
x  
3x  5
5 . lim
6x  8
x  
4x  x

x 5x
6
3

2
2 . lim
x  
2x  5
3
5x  2x  1
3
3 . lim
x  
2
1  3x
x 2
2
4 . lim
x  
6.
3x  6
lim 
x  
x  5x  x
6
3
3

EXERCISES:
A. Evaluate the following limits.

1 . lim 4 x  5 x  2
x 3
2

6. lim
x
1
3x  1
9x  1
2
3
2 . lim
x  1
2x  1
2x  3x  2x  3
3
7. lim
x  3x  4
2
x 1
2
x1
x  3x  4
2
3 . lim
x 1
3
x  1
2
8. lim
 y  1  y 2
 2y  3
y1
y  2y 1


2
1
 4y3  8y
4 . lim 
y 2
 y4
x 8
3



9. lim 2 x  9 x  19
x 5
3
5 . lim
x 2
x2
4
3
w 7w 7
2
10. lim
w  1
w  4w  5
2

1
2

EXERCISES:
B. Sketch the graph of the following functions
and the indicated limit if it exists. find
.
4  x
1. f ( x )  
x  4
a . lim
x  4

f(x)
x1
x  -4
if
x -4
b . lim
2 x  3

2. g ( x )  2
7 - 2x

a . lim  g(x)
if
x  4

f(x) c . lim f ( x )
x  4
if
x 1
if
x 1
if
x 1
b . lim  g(x)
x1
c . lim g ( x )
x1
x
3. f ( x )  
3
a . lim  f(x)
x0
4. g ( x ) 
x 4
1
2
if
x 0
b . lim  f(x) c . lim f ( x )
x0
x0
b . lim  f(x)
x 4
c . lim f ( x )
x 4
2x  1
a . lim  f(x)
x
x 0
4 x
a . lim  f(x)
5. g ( x ) 
if
.
b . lim  f(x)
x
1
2
c . lim f ( x )
x
1
2