Section 1.3: Evaluating Limits
Analytically
Example
Let f
. Find:
4

5
4

2and g

357
2
357
2
4
157
4
8
4
2
3 g47
f 4
2
Direct Substitution
One of the easiest and most useful ways to evaluate a limit
analytically is direct substitution (substitution and
evaluation):
If you can plug c into f(x) and generate a real
number answer in the range of f(x), that generally
implies that the limit exists (assuming f(x) is
continuous at c).
Example:
limx 2 8
3
Always check for
substitution first.
3
x
2
The slides that follow investigate why Direct Substitution is valid.
Properties of Limits
Let b and c be real numbers, let n be a positive integer, and
let f and g be functions with the following limits:
l
i
m
f()
x
L
x

c
l
i
m
g
()
x
K
x

c
Constant Function
lim
bb
Limit of x
lim
xc
Limit of a Power of x
Scalar Multiple
x
c
x
c
lim
x
c
n
n
x

c
l
i
m
b
f
(
x
)

b

L


x

c
Properties of Limits
Let b and c be real numbers, let n be a positive integer, and
let f and g be functions with the following limits:
l
i
m
f()
x
L
x

c
Sum Difference
Product
Quotient
Power
l
i
m
g
()
x
K
x

c
l
i
m
f
(
x
)(

g
x
)

L

K


x

c
l
i
m
f
(
x
)

g
(
xL
)


K


x

c
fx
() L
l
i
m , K

0
xc

g
(
x
) K
l
i
m
()
L
fx

n
x

c
n
Example
Let lim f  x   4 and lim g  x   2 . Find the following
x 5
x 5
limits.
.lim
f
x

g
x
1
.lim
f
x

5
g
x
. lim









3

2
f
x
x

x

5g
x

5
x

5
lim
f
x
+
lim
5
g
x
f
x

lim
g
x



 lim





x

5
x

5
x

5
f
x
+
5
lim
g
x





 lim
x

5


x

5
4
+
5

2


6

x

5
lim gx
x5
4
2


8
lim f x
x5
4
2


2
Example 2
E
v
a
l
u
a
t
e
l
i
m
2
x

9
x

3
x

1
1


532
x

2
5
3
2

l
i
m
2
x

l
i
m
9
x

l
i
m
3
x

l
i
m
1
1






x

2
x

2
x

2
x

2
Sum/Difference Property
Direct Substitution
5 
3 
2


2
l
i
m
x

9
l
i
m
x

3
l
i
m
x

1
1
x

2
x

2
x

2



Multiple and Constant Properties
5
3
2




2
l
i
m
x

9
l
i
m
x

3
l
i
m
x

1
1
x

2
x

2
x

2




2
2
2

3
2

1
1

9



5
3
2
Power Property
Limit of x Property
7
Direct Substitution
Direct substitution is a valid analytical method to evaluate
the following limits.
•
If p is a polynomial function and c is a real number, then:
l
i
m
p
(
x
)

p
(
c
)
xc

•
If r is a rational function given by r(x) = p(x)/q(x), and c is a real
number, then
p
(
c
)
l
i
m
r
(
x
)

r
(
c
)
,q
(
c
)

0
x

c
q
(
c
)
•
If a radical function where n is a positive integer. The following
limit is valid for all c if n is odd and only c>0 when n is even:
n
n
lim
x
c
x

c
Direct Substitution
Direct substitution is a valid analytical method to evaluate
the following limits.
•
i
m
g
(
x
)

L
a
n
d
l
i
m
f
(
x
)

f
(
L
)
If the f and g are functions such that l
x

c
x

L
Then the limit of the composition is:
 
l
i
m
f
(
g
(
x
)
)

f
l
i
m
g
(
x
)

f
(
L
)
x

c
•
x

c
If c is a real number in the domain of a trigonometric function
then:
l
i
m
s
i
n
x

s
i
n
c
l
i
m
c
o
s
x

c
o
s
c
l
i
m
t
a
n
x

t
a
n
c
l
i
m
c
o
tx

c
o
t
c
l
i
m
s
e
c
x

s
e
c
c
l
i
m
c
s
c
x

c
s
c
c
xc

xc

xc

xc

xc

xc

Example
x

2


E
v
a
l
u
a
t
el
i
m
 
x

3x
6


3 2

3 6
1

9
Direct Substitution can
be used since the
function is well defined
at x=3
For what value(s) of x can the limit not be
evaluated using direct substitution?
At x=-6 since it makes the
denominator 0:

66

0
Indeterminate Form
Evaluate the limit analytically:
x
4x
4
2
x
x2
x
2 
2
lim
0
 2222 
 
0
2242
 4
An example of an
indeterminate form
because the limit can
not be determined. 1/0
is another example.
Often limits can not be evaluated at a value using
Direct Substitution. If this is the case, try to find
another function that agrees with the original
function except at the point in question. In other
words…
How can we simplify:
x24x4
x2x2
?
Strategies for Finding Limits
To find limits analytically, try the following:
1. Direct Substitution (Try this FIRST)
2. If Direct Substitution fails, then rewrite then find a
function that is equivalent to the original function except
at one point. Then use Direct Substitution. Methods
for this include…
•
•
•
•
•
Factoring/Dividing Out Technique
Rationalize Numerator/Denominator
Eliminating Embedded Denominators
Trigonometric Identities
Legal Creativity
Example 1
Evaluate the limit analytically:
x2
4x
4
2
x
x2
x
2 
lim
At first Direct
Substitution fails
because x=2 results
in dividing by zero

2

2
x
x


lim

2

1
x

x

2x

2

2
x
x


lim

2

1
x

x

2x
lim
x
2

22
21
0
x2
x
1
Factor the numerator and
denominator
Cancel common factors
This function is
equivalent to the
original function except
at x=2
Direct substitution
Example 2
Evaluate the limit analytically:
y22
y2
lim
y
2

y22
Rationalize the numerator
y22

24
y


l
i
m

2

2

2
y
 y

y

2
y

2

l
i
m

2
2

2
y

 y

y

2
1

lim
y

22
y

2
1
2

2

2


1
4
Cancel common factors
Direct substitution
Example 3
Evaluate the limit analytically:
11
x 3
3

x
lim
limx3  33 xx 
3
x
x
3
x3
Cancel the
denominators
of the fractions
in the
numerator
x

3

x3
lim
x33x
If the subtraction is
backwards, Factoring a
negative 1 to flip the signs
lim
x33x
Cancel common factors
x
3

x3
x
3
1
3x
x
3

1
1
3
3
9
lim


Direct substitution
Example 4
Evaluate the2limit analytically:
h 5 h 5 25
h 5 25
h
h
h

0
h0
h 2 10h 2525
Expand the
h
the expression
h0 2
to see if
h 10h
anything
h
h

0
cancels
h  h 10
h
h0
lim
 lim
 lim
 lim



 lim
 lim h 10
h0
 0 10
 10
Factor to see if anything
cancels
Direct substitution
Example 5
Evaluate the limit analytically:
sin x
1
cos x
1tan x
cosx
sin x cosx cosx
sin x cosx
x  4
x  4
cosx sin x
Rewrite the
sin x cosx  cosx
tangent
x  4 
function using
sin x cosx 
cosine and
sine
x  4 sin x cosx  cosx
1
cosx
x  4
1
cos4 
1
2 2
lim
 lim

Eliminate the
embedded fraction
 lim



 lim

 lim


 2
If the subtraction
is backwards,
Factoring a
negative 1 to flip
the signs
Direct substitution
Two “Freebie” Limits
The following limits can be assumed to be true (they will be
proven later in the year) to assist in finding other limits:
sinx
lim
1
x
0 x
1c
o
sx
lim

0
x

0
x
Use the identities to help with these limits. They are
located on the first page of your textbook.
Example
Evaluate the limit analytically:
lim
x 0
sin 3 x
5x
If 3x is the
input of the
sine function
then 3x needs
to be in the
denominator

3
3
3x
 lim 3sin
53 x
x 0
 lim 53  sin3 x3 x
x 0
3
sin 3 x
5
3x
x0
  lim
 1

3
5
3
5
Isolate the “freebie”
Scalar Multiple Property
Assumed Trig Limit
Example
Evaluate the limit analytically:
1cosx 1cosx
1cosx cosx cos2 x
x sin x 1cosx  lim
x sin x1cosx 
x

0
x0
2

lim
Try multiplying
by the
reciprocal





 lim
1cos x
x sin x1cosx
 lim
sin 2 x
x sin x1cosx
x0
x0
 lim
Use the
Trigonometry
Laws
sin x
x 1cosx 
x0 
sin x
1
Split up the limits
x
1cosx
x0
x0
1
1cos0
A freebie limit and Direct
substitution
1
2
 lim
1

 lim