Consider the rational function
f(x)=
x
2
−6x−7
x−7
f(x)=
x
2
−6x−7
x−7
The function can be factored as follows:
f(x)=
(
x−7
)
(
x+1
)
x−7
,
which gives us
f(x)=x+1,x≠7.
f(x)=
(
x−7
)
(
x+1
)
x−7
,
which gives us
f(x)=x+1,x≠7.
Does this mean the function
f
f is the same as the function
g(x)=x+1?
g(x)=x+1?
The answer is no. Function
f
f does not have
x=7
x=7
in its domain, but
g
g does. Graphically, we observe there is a hole in the graph of
f(
x
)
f(
x
)
at
x=7,
x=7,
as shown in Figure 1 and no such hole in the graph of
g(
x
),
g(
x
),
as shown in Figure 2.
So, do these two different functions also have different limits as
x
x approaches 7?
Not necessarily. Remember, in determining a limit of a function as
x
x approaches
a,
a,
what matters is whether the output approaches a real number as we get close to
x=a.
x=a.
The existence of a limit does not depend on what happens when
x
x equals
a.
a.
Look again at Figure 1 and Figure 2. Notice that in both graphs, as
x
x approaches 7, the output values approach 8. This means
lim
x→7
f(x)=
lim
x→7
g(x).
lim
x→7
f(x)=
lim
x→7
g(x).
Remember that when determining a limit, the concern is what occurs near
x=a,
x=a,
not at
x=a.
x=a.
In this section, we will use a variety of methods, such as rewriting functions by factoring, to evaluate the limit. These methods will give us formal verification for what we formerly accomplished by intuition.
Graphing a function or exploring a table of values to determine a limit can be cumbersome and timeconsuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits.
Knowing the properties of limits allows us to compute limits directly. We can add, subtract, multiply, and divide the limits of functions as if we were performing the operations on the functions themselves to find the limit of the result. Similarly, we can find the limit of a function raised to a power by raising the limit to that power. We can also find the limit of the root of a function by taking the root of the limit. Using these operations on limits, we can find the limits of more complex functions by finding the limits of their simpler component functions.
Let
a, k, A,
a, k, A,
and
B
B represent real numbers, and
f
f and
g
g be functions, such that
lim
x→a
f(x)=A
lim
x→a
f(x)=A
and
lim
x→a
g(x)=B.
lim
x→a
g(x)=B.
For limits that exist and are finite, the properties of limits are summarized in Table 1
Table 1
Constant, k 
lim
x→a
k=k
lim
x→a
k=k

Constant times a function 
lim
x→a
[
k⋅f(x)
]=k
lim
x→a
f(x)=kA
lim
x→a
[
k⋅f(x)
]=k
lim
x→a
f(x)=kA

Sum of functions 
lim
x→a
[
f(x)+g(x)
]=
lim
x→a
f(x)+
lim
x→a
g(x)=A+B
lim
x→a
[
f(x)+g(x)
]=
lim
x→a
f(x)+
lim
x→a
g(x)=A+B

Difference of functions 
lim
x→a
[
f(x)−g(x)
]=
lim
x→a
f(x)−
lim
x→a
g(x)=A−B
lim
x→a
[
f(x)−g(x)
]=
lim
x→a
f(x)−
lim
x→a
g(x)=A−B

Product of functions 
lim
x→a
[
f(x)⋅g(x)
]=
lim
x→a
f(x)⋅
lim
x→a
g(x)=A⋅B
lim
x→a
[
f(x)⋅g(x)
]=
lim
x→a
f(x)⋅
lim
x→a
g(x)=A⋅B

Quotient of functions 
lim
x→a
f(x)
g(x)
=
lim
x→a
f(x)
lim
x→a
g(x)
=
A
B
,B≠0
lim
x→a
f(x)
g(x)
=
lim
x→a
f(x)
lim
x→a
g(x)
=
A
B
,B≠0

Function raised to an exponent 
lim
x→a
[f(x)]
n
=
[
lim
x→∞
f(x)
]
n
=
A
n
,
lim
x→a
[f(x)]
n
=
[
lim
x→∞
f(x)
]
n
=
A
n
,
where
n
n
is a positive integer 
nth root of a function, where n is a positive integer 
lim
x→a
f(x)
n
=
lim
x→a
[
f(x)
]
n
=
A
n
lim
x→a
f(x)
n
=
lim
x→a
[
f(x)
]
n
=
A
n

Polynomial function 
lim
x→a
p(x)=p(a)
lim
x→a
p(x)=p(a)

Evaluate
lim
x→3
(
2x+5
).
lim
x→3
(
2x+5
).
lim
x→3
(2x+5)=
lim
x→3
(2x)+
lim
x→3
(5)
Sum of functions property
=
2lim
x→3
(x)+
lim
x→3
(5)
Constant times a function property
=2(3)+5
Evaluate
=11
lim
x→3
(2x+5)=
lim
x→3
(2x)+
lim
x→3
(5)
Sum of functions property
=
2lim
x→3
(x)+
lim
x→3
(5)
Constant times a function property
=2(3)+5
Evaluate
=11
Evaluate the following limit:
lim
x→−12
(
−2x+2
).
lim
x→−12
(
−2x+2
).
Not all functions or their limits involve simple addition, subtraction, or multiplication. Some may include polynomials. Recall that a polynomial is an expression consisting of the sum of two or more terms, each of which consists of a constant and a variable raised to a nonnegative integral power. To find the limit of a polynomial function, we can find the limits of the individual terms of the function, and then add them together. Also, the limit of a polynomial function as
x
x approaches
a
a is equivalent to simply evaluating the function for
a
a.
Given a function containing a polynomial, find its limit.
 Use the properties of limits to break up the polynomial into individual terms.
 Find the limits of the individual terms.
 Add the limits together.
 Alternatively, evaluate the function for
a
a.
Evaluate
lim
x→3
(
5
x
2
).
lim
x→3
(
5
x
2
).
lim
x→3
(5
x
2
)=5
lim
x→3
(
x
2
)
Constant times a function property
=5(
3
2
)
Function raised to an exponent property
=45
lim
x→3
(5
x
2
)=5
lim
x→3
(
x
2
)
Constant times a function property
=5(
3
2
)
Function raised to an exponent property
=45
Evaluate
lim
x→4
(
x
3
−5).
lim
x→4
(
x
3
−5).
Evaluate
lim
x→5
(
2
x
3
−3x+1
).
lim
x→5
(
2
x
3
−3x+1
).
lim
x→5
(2
x
3
−3x+1)=
lim
x→5
(2
x
3
)−
lim
x→5
(3x)+
lim
x→5
(1)
Sum of functions
=
2lim
x→5
(
x
3
)−
3lim
x→5
(x)+
lim
x→5
(1)
Constant times a function
=2(
5
3
)−3(5)+1
Function raised to an exponent
=236
Evaluate
lim
x→5
(2
x
3
−3x+1)=
lim
x→5
(2
x
3
)−
lim
x→5
(3x)+
lim
x→5
(1)
Sum of functions
=
2lim
x→5
(
x
3
)−
3lim
x→5
(x)+
lim
x→5
(1)
Constant times a function
=2(
5
3
)−3(5)+1
Function raised to an exponent
=236
Evaluate
Evaluate the following limit:
lim
x→−1
(
x
4
−4
x
3
+5
).
lim
x→−1
(
x
4
−4
x
3
+5
).
When a limit includes a power or a root, we need another property to help us evaluate it. The square of the limit of a function equals the limit of the square of the function; the same goes for higher powers. Likewise, the square root of the limit of a function equals the limit of the square root of the function; the same holds true for higher roots.
Evaluate
lim
x→2
(
3x+1
)
5
.
lim
x→2
(
3x+1
)
5
.
We will take the limit of the function as
x
x approaches 2 and raise the result to the 5^{th} power.
lim
x→2
(3x+1)
5
=
(
lim
x→2
(3x+1))
5
=
(3(2)+1)
5
=
7
5
=16,807
lim
x→2
(3x+1)
5
=
(
lim
x→2
(3x+1))
5
=
(3(2)+1)
5
=
7
5
=16,807
Evaluate the following limit:
lim
x→−4
(
10x+36
)
3
.
lim
x→−4
(
10x+36
)
3
.
If we can’t directly apply the properties of a limit, for example in
lim
x→2
(
x
2
+6x+8
x−2
)
lim
x→2
(
x
2
+6x+8
x−2
), can we still determine the limit of the function as
x
x approaches
a
a?
Yes. Some functions may be algebraically rearranged so that one can evaluate the limit of a simplified equivalent form of the function.
Finding the limit of a function expressed as a quotient can be more complicated. We often need to rewrite the function algebraically before applying the properties of a limit. If the denominator evaluates to 0 when we apply the properties of a limit directly, we must rewrite the quotient in a different form. One approach is to write the quotient in factored form and simplify.
Given the limit of a function in quotient form, use factoring to evaluate it.
 Factor the numerator and denominator completely.
 Simplify by dividing any factors common to the numerator and denominator.
 Evaluate the resulting limit, remembering to use the correct domain.
Evaluate
lim
x→2
(
x
2
−6x+8
x−2
).
lim
x→2
(
x
2
−6x+8
x−2
).
Factor where possible, and simplify.
lim
x→2
(
x
2
−6x+8
x−2
)=
lim
x→2
(
(x−2)(x−4)
x−2
)
Factor the numerator.
=
lim
x→2
(
(x−2)
(x−4)
x−2
)
Cancel the common factors.
=
lim
x→2
(x−4)
Evaluate.
=2−4=−2
lim
x→2
(
x
2
−6x+8
x−2
)=
lim
x→2
(
(x−2)(x−4)
x−2
)
Factor the numerator.
=
lim
x→2
(
(x−2)
(x−4)
x−2
)
Cancel the common factors.
=
lim
x→2
(x−4)
Evaluate.
=2−4=−2
Evaluate the following limit:
lim
x→7
(
x
2
−11x+28
7−x
).
lim
x→7
(
x
2
−11x+28
7−x
).
Evaluate
lim
x→5
(
1
x
−
1
5
x−5
).
lim
x→5
(
1
x
−
1
5
x−5
).
Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator.
Evaluate
lim
x→−5
(
1
5
+
1
x
10+2x
).
lim
x→−5
(
1
5
+
1
x
10+2x
).
Given a limit of a function containing a root, use a conjugate to evaluate.
 If the quotient as given is not in indeterminate
(
0
0
)
(
0
0
)
form, evaluate directly.
 Otherwise, rewrite the sum (or difference) of two quotients as a single quotient, using the least common denominator (LCD).
 If the numerator includes a root, rationalize the numerator; multiply the numerator and denominator by the conjugate of the numerator. Recall that
a±
b
a±
b
are conjugates.
 Simplify.
 Evaluate the resulting limit.
Evaluate
lim
x→0
(
25−x
−5
x
).
lim
x→0
(
25−x
−5
x
).
lim
x→0
(
25−x
−5
x
)=
lim
x→0
(
(
25−x
−5
)
x
⋅
(
25−x
+5
)
(
25−x
+5
)
)
Multiply numerator and denominator by the conjugate.
=
lim
x→0
(
(
25−x
)−25
x(
25−x
+5
)
)
Multiply: (
25−x
−5
)⋅(
25−x
+5
)=(
25−x
)−25.
=
lim
x→0
(
−x
x(
25−x
+5
)
)
Combine like terms.
=
lim
x→0
(
−
x
x
(
25−x
+5
)
)
Simplify
−x
x
=−1.
=
−1
25−0
+5
Evaluate.
=
−1
5+5
=−
1
10
lim
x→0
(
25−x
−5
x
)=
lim
x→0
(
(
25−x
−5
)
x
⋅
(
25−x
+5
)
(
25−x
+5
)
)
Multiply numerator and denominator by the conjugate.
=
lim
x→0
(
(
25−x
)−25
x(
25−x
+5
)
)
Multiply: (
25−x
−5
)⋅(
25−x
+5
)=(
25−x
)−25.
=
lim
x→0
(
−x
x(
25−x
+5
)
)
Combine like terms.
=
lim
x→0
(
−
x
x
(
25−x
+5
)
)
Simplify
−x
x
=−1.
=
−1
25−0
+5
Evaluate.
=
−1
5+5
=−
1
10
Evaluate the following limit:
lim
h→0
(
16−h
−4
h
).
lim
h→0
(
16−h
−4
h
).
Evaluate
lim
x→4
(
4−x
x
−2
).
lim
x→4
(
4−x
x
−2
).
lim
x→4
(
4−x
x
−2
) =
lim
x→4
(
(2+
x
)(2−
x
)
x
−2
)
Factor.
=
lim
x→4
(
(2+
x
)
(2−
x
)
−
(2−
x
)
)
Factor −1 out of the denominator. Simplify.
=
lim
x→4
−(2+
x
)
Evaluate.
=−(2+
4
)
=−4
lim
x→4
(
4−x
x
−2
) =
lim
x→4
(
(2+
x
)(2−
x
)
x
−2
)
Factor.
=
lim
x→4
(
(2+
x
)
(2−
x
)
−
(2−
x
)
)
Factor −1 out of the denominator. Simplify.
=
lim
x→4
−(2+
x
)
Evaluate.
=−(2+
4
)
=−4
Evaluate the following limit:
lim
x→3
(
x−3
x
−
3
).
lim
x→3
(
x−3
x
−
3
).
Given a quotient with absolute values, evaluate its limit.
 Try factoring or finding the LCD.
 If the limit cannot be found, choose several values close to and on either side of the input where the function is undefined.
 Use the numeric evidence to estimate the limits on both sides.
Evaluate
lim
x→7

x−7

x−7
.
lim
x→7

x−7

x−7
.
The function is undefined at
x=7
,
x=7
,so we will try values close to 7 from the left and the right.
Lefthand limit:

6.9−7

6.9−7
=

6.99−7

6.99−7
=

6.999−7

6.999−7
=−1

6.9−7

6.9−7
=

6.99−7

6.99−7
=

6.999−7

6.999−7
=−1
Righthand limit:

7.1−7

7.1−7
=

7.01−7

7.01−7
=

7.001−7

7.001−7
=1

7.1−7

7.1−7
=

7.01−7

7.01−7
=

7.001−7

7.001−7
=1
Since the left and righthand limits are not equal, there is no limit.
Evaluate
lim
x→
6
+
6−x

x−6

.
lim
x→
6
+
6−x

x−6

.
Access the following online resource for additional instruction and practice with properties of limits.
 The properties of limits can be used to perform operations on the limits of functions rather than the functions themselves. See Example 1.
 The limit of a polynomial function can be found by finding the sum of the limits of the individual terms. See Example 2 and Example 3.
 The limit of a function that has been raised to a power equals the same power of the limit of the function. Another method is direct substitution. See Example 4.
 The limit of the root of a function equals the corresponding root of the limit of the function.
 One way to find the limit of a function expressed as a quotient is to write the quotient in factored form and simplify. See Example 5.
 Another method of finding the limit of a complex fraction is to find the LCD. See Example 6.
 A limit containing a function containing a root may be evaluated using a conjugate. See Example 7.
 The limits of some functions expressed as quotients can be found by factoring. See Example 8.
 One way to evaluate the limit of a quotient containing absolute values is by using numeric evidence. Setting it up piecewise can also be useful. See Example 9.
Give an example of a type of function
f
f whose limit, as
x
x approaches
a,
a,
is
f(
a
).
f(
a
).
If
f
f is a polynomial function, the limit of a polynomial function as
x
x approaches
a
a will always be
f(
a
).
f(
a
).
When direct substitution is used to evaluate the limit of a rational function as
x
x approaches
a
a and the result is
f(
a
)=
0
0
,
f(
a
)=
0
0
,
does this mean that the limit of
f
f does not exist?
What does it mean to say the limit of
f(
x
)
,
f(
x
)
, as
x
x approaches
c
, c, is undefined?
It could mean either (1) the values of the function increase or decrease without bound as
x
x approaches
c,
c,
or (2) the left and righthand limits are not equal.
For the following exercises, evaluate the limits algebraically.
lim
x→0
(
3
)
lim
x→0
(
3
)
lim
x→2
(
−5x
x
2
−1
)
lim
x→2
(
−5x
x
2
−1
)
lim
x→2
(
x
2
−5x+6
x+2
)
lim
x→2
(
x
2
−5x+6
x+2
)
lim
x→3
(
x
2
−9
x−3
)
lim
x→3
(
x
2
−9
x−3
)
lim
x→−1
(
x
2
−2x−3
x+1
)
lim
x→−1
(
x
2
−2x−3
x+1
)
lim
x→
3
2
(
6
x
2
−17x+12
2x−3
)
lim
x→
3
2
(
6
x
2
−17x+12
2x−3
)
lim
x→−
7
2
(
8
x
2
+18x−35
2x+7
)
lim
x→−
7
2
(
8
x
2
+18x−35
2x+7
)
lim
x→3
(
x
2
−9
x−5x+6
)
lim
x→3
(
x
2
−9
x−5x+6
)
lim
x→−3
(
−7
x
4
−21
x
3
−12
x
4
+108
x
2
)
lim
x→−3
(
−7
x
4
−21
x
3
−12
x
4
+108
x
2
)
lim
x→3
(
x
2
+2x−3
x−3
)
lim
x→3
(
x
2
+2x−3
x−3
)
lim
h→0
(
(
3+h
)
3
−27
h
)
lim
h→0
(
(
3+h
)
3
−27
h
)
lim
h→0
(
(
2−h
)
3
−8
h
)
lim
h→0
(
(
2−h
)
3
−8
h
)
lim
h→0
(
(
h+3
)
2
−9
h
)
lim
h→0
(
(
h+3
)
2
−9
h
)
lim
h→0
(
5−h
−
5
h
)
lim
h→0
(
5−h
−
5
h
)
lim
x→0
(
3−x
−
3
x
)
lim
x→0
(
3−x
−
3
x
)
lim
x→9
(
x
2
−81
3−
x
)
lim
x→9
(
x
2
−81
3−
x
)
lim
x→1
(
x
−
x
2
1−
x
)
lim
x→1
(
x
−
x
2
1−
x
)
lim
x→0
(
x
1+2x
−1
)
lim
x→0
(
x
1+2x
−1
)
lim
x→
1
2
(
x
2
−
1
4
2x−1
)
lim
x→
1
2
(
x
2
−
1
4
2x−1
)
lim
x→4
(
x
3
−64
x
2
−16
)
lim
x→4
(
x
3
−64
x
2
−16
)
lim
x→
2
−
(
x−2
x−2
)
lim
x→
2
−
(
x−2
x−2
)
lim
x→
2
+
(

x−2

x−2
)
lim
x→
2
+
(

x−2

x−2
)
lim
x→2
(

x−2

x−2
)
lim
x→2
(

x−2

x−2
)
lim
x→
4
−
(

x−4

4−x
)
lim
x→
4
−
(

x−4

4−x
)
lim
x→
4
+
(

x−4

4−x
)
lim
x→
4
+
(

x−4

4−x
)
lim
x→4
(

x−4

4−x
)
lim
x→4
(

x−4

4−x
)
lim
x→2
(
−8+6x−
x
2
x−2
)
lim
x→2
(
−8+6x−
x
2
x−2
)
For the following exercise, use the given information to evaluate the limits:
lim
x→c
f(x)=3,
lim
x→c
f(x)=3,
lim
x→c
g(
x
)=5
lim
x→c
g(
x
)=5
lim
x→c
[
2f(x)+
g(x)
]
lim
x→c
[
2f(x)+
g(x)
]
lim
x→c
[
3f(x)+
g(x)
]
lim
x→c
[
3f(x)+
g(x)
]
lim
x→c
f(x)
g(x)
lim
x→c
f(x)
g(x)
For the following exercises, evaluate the following limits.
lim
x→2
cos(
πx
)
lim
x→2
cos(
πx
)
lim
x→2
sin(
πx
)
lim
x→2
sin(
πx
)
lim
x→2
sin(
π
x
)
lim
x→2
sin(
π
x
)
f(x)={
2
x
2
+2x+1,
x≤0
x−3,
x>0
;
lim
x→
0
+
f(x)
f(x)={
2
x
2
+2x+1,
x≤0
x−3,
x>0
;
lim
x→
0
+
f(x)
f(x)={
2
x
2
+2x+1,
x≤0
x−3,
x>0
;
lim
x→
0
−
f(x)
f(x)={
2
x
2
+2x+1,
x≤0
x−3,
x>0
;
lim
x→
0
−
f(x)
f(x)={
2
x
2
+2x+1,
x≤0
x−3,
x>0
;
lim
x→0
f(x)
f(x)={
2
x
2
+2x+1,
x≤0
x−3,
x>0
;
lim
x→0
f(x)
does not exist; righthand limit is not the same as the lefthand limit.
lim
x→4
x+5
−3
x−4
lim
x→4
x+5
−3
x−4
lim
x→
2
+
(2x−〚x〛)
lim
x→
2
+
(2x−〚x〛)
lim
x→2
x+7
−3
x
2
−x−2
lim
x→2
x+7
−3
x
2
−x−2
lim
x→
3
+
x
2
x
2
−9
lim
x→
3
+
x
2
x
2
−9
Limit does not exist; limit approaches infinity.
For the following exercises, find the average rate of change
f(x+h)−f(x)
h
.
f(x+h)−f(x)
h
.
f(x)=2
x
2
−1
f(x)=2
x
2
−1
f(x)=
x
2
+3x+4
f(x)=
x
2
+3x+4
f(x)=
x
2
+4x−100
f(x)=
x
2
+4x−100
f(x)=3
x
2
+1
f(x)=3
x
2
+1
cos(x+h)−cos(x)
h
cos(x+h)−cos(x)
h
f(x)=2
x
3
−4x
f(x)=2
x
3
−4x
Find an equation that could be represented by Figure 3.
Find an equation that could be represented by Figure 4.
f(
x
)=
x
2
+5x+6
x+3
f(
x
)=
x
2
+5x+6
x+3
For the following exercises, refer to Figure 5.
What is the righthand limit of the function as
x
x approaches 0?
What is the lefthand limit of the function as
x
x approaches 0?
The position function
s(t)=−16
t
2
+144t
s(t)=−16
t
2
+144t
gives the position of a projectile as a function of time. Find the average velocity (average rate of change) on the interval
[
1,2
]
[
1,2
]
.
The height of a projectile is given by
s(t)=−64
t
2
+192t
s(t)=−64
t
2
+192t
Find the average rate of change of the height from
t=1
t=1
second to
t=1.5
t=1.5
seconds.
The amount of money in an account after
t
t years compounded continuously at 4.25% interest is given by the formula
A=
A
0
e
0.0425t
,
A=
A
0
e
0.0425t
,where
A
0
A
0
is the initial amount invested. Find the average rate of change of the balance of the account from
t=1
t=1
year to
t=2
t=2
years if the initial amount invested is $1,000.00.
 properties of limits:
a collection of theorems for finding limits of functions by performing mathematical operations on the limits
Analysis
When the limit of a rational function cannot be evaluated directly, factored forms of the numerator and denominator may simplify to a result that can be evaluated.
Notice, the function
is equivalent to the function
Notice that the limit exists even though the function is not defined at
x = 2 .
x = 2 .