Computing limits
This is an outline of some ways to evaluate
lim f ( x )
x a
or to determine it does not exist.
Case 1: f(x) is given by an algebraic rule and f(a) is defined then f(a) is the limit. In other
words, if substituting a for x gives a meaningful answer then that is the limit.
Examples:
x 1
lim
x0
x2

1
lim ( x
2
x 2
2
 3 x )  10
x3
lim
x  3
Examples of non-meaningful expressions are
0
x
0
0
,
1
,
0
0
0
and there are others that we
will discuss if they come up.
nonzero
Substitution gives
0
. In this case the limit does not exist as a number, the
magnitude of the function goes to infinity.
0
If Substitution gives
0
. This is indeterminate, the limit could be any number or not
exist. Do Algebra.
p( x)
Case 2: Case 2: f(x) is a rational function,
nonzero
substitution gives
q( x)
and q(a) is 0 but p(a) is not 0. Then
so the limit does not exist.
0
Case 3: f(x) is a rational function,
p( x)
q( x)
0
, and substitution gives
.
0
This means both p(a) and q(a) have (x-a) as a factor. Factor the numerator and
denominator and cancel common factors. Substitute again and re-evaluate it.
Example:
x
x
2
2
9
x6
x
lim
x 3

x
2
2
0
9
Substitution of 3 for x gives
x6
( x  3 )( x  3 )
( x  3 )( x  2 )

x3
x2
if
x  3.
taking the limit. Now substitution gives
Example:
x
x
2
2
 2x
 4x  4
x
lim
x  2

x
2
2
 4x  4
( x  2)
does not exist.
2

5
.
.
This is OK since we do not let x equal 3 in
lim f ( x ) 
x3
6
5
.
0
 2x
x( x  2)
6
0
x
x2
Substitution of -2 for x gives
if
x  2
0
.
Now substituting x= - 2 gives
2
0
, so the limit
Case 4: Conjugating is needed to simplify. Recall that
xa ( x 
a )(
x 
x 
and
a)
a 
(
x 
c
2
d
2
a )(
x 
x 
a
 ( c  d )( c  d )
a)

xa
x 
so also
. This is
a
called conjugating.
Example: Find
lim
x1
x 1
x 1
lim
x1
 lim
x1 (x
x 1
x 1
0
Substitution gives
x 1
 1)(
x  1)
1
 lim
x1
x 1

0
, so we conjugate and cancel.
1
2
Case 5: The above do not fit. Look at the graph or try to use the squeeze theorem for
some possible strategies. Find
lim sin(
x0
1
x
)
if it exists. Look at the graph and you will see
that the function oscillates between -1 and 1 and takes on all values in between no matter
how close we get to x=0. The limit does not exist.
Find
1
lim x sin  
x0
x
1
 1  sin    1
x
so
1
 x  sin    x
x
-x and x both approach 0 so the
limit is 0. (The function is squeezed between two functions which have the same limit.)