Computing Limits: Part II
In this handout, we will solve limits for continuous and non-continuous functions.
1. Indeterminate Case
.
Let’s look at an example to illustrate this situation.
Example 1:
Evaluate
Solution: In the Introduction to Limits handout we learned that for all continuous
( )
functions
( )
Can we apply this property to our function in Example 1? Is f(x) continuous?
Figure 1: ( )
In this example, if we try to plug
in
, we end up with .
Looking at Figure 1 we see that ( )
has a hole at
and therefore is
not a continuous at that point.
Knowing this, can we still solve for
?
YES, we can! Remember the definition of a limit asks us to determine the
value of f(x) as x approaches (but does not necessarily equal) a.
Thus, when
is substituted into a function and the result is we have the
"Indeterminate" Case.
When working with the indeterminate case our goal is to get rid of the zero in the
denominator by factoring.
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Computing Limits: Part II
Looking at the function,
we notice the top function can be factored as a difference
of squares.
Once the top and bottom functions are factored out our goal is to reduce the function as
much as possible.
(
Since (
)(
)
) is common to both the numerator and denominator we can eliminate it.
Thus, we are left to solve
Since ( ) =
solve for the limit.
is a continuous at
(
Therefore,
we can substitute it into the function and
)
Looking at Figure 1, determine if we have correctly solved for the limit.
Example 2:
Evaluate
Solution: Once again, we see that if we plug in
, we will end up with
. Notice
that we can factor both the numerator and denominator of the function.
(
(
)(
)(
)
)
Thus, we are left to solve
Since ( ) =
is continuous at
we can plug it into the function and solve for
the limit.
Therefore,
( )
( )
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Computing Limits: Part II
Example 3:
√
Evaluate
Solution: Again, in this example we see that if we plug in
we will get
.
In order to get rid of the zero in the denominator let’s ‘RATIONALIZE’ the function by
multiplying the numerator and denominator by
√
√
(
√
)
(
√
)
(
√
)
(
√
)
.
(
(
)(
)
√
)
(
)(
√
)
)(
) Moreover; we can
Looking at the numerator we can factor
to (
get rid of the(
) factor in the numerator and the denominator.
(
(
Since ( ) =
√
)(
)
√
is continuous at
(
)(
)(
)
√
)
√
we can plug it into the function and solve
for the limit.
Therefore,
√
√
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Computing Limits: Part II
2. Evaluating Limits at Asymptotes
Let’s look at an example to illustrate this situation.
Example 1:
Evaluate
Solution: Is ( ) continuous at
?
Figure 2: ( )
In this example, if we plug in
, we end
up with .
Looking at Figure 2 we see that ( ) fails to
be continuous at the point
.
In order to solve for the limit, we learned that
sometimes we can eliminate the zero in the
denominator with some algebraic
manipulations.
However; in this case we cannot.
Thus, when the zero in the denominator cannot be eliminated the resulting limit will be
either
. When this happens, we say that we have a vertical asymptote at that
point.
To understand why this is true, let’s take a look at Figure 2. Take the limit as x
approaches 0 from the right hand side. What does ( ) approach? ( )
approaches
. If we take the limit from the left hand side of 0, ( ) approaches - .
Since the right hand side and the left hand side of the limit do not equal, the limit does
not exist.
Furthermore, we say that there is an asymptote at
.
Remark: Asymptotes are points in the denominator of RATIONAL FUNCTIONS where
evaluating at that point would result in a number divided by 0.
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Computing Limits: Part II
3. Evaluating Limits at Infinity
We’ve now seen how evaluating limits as we approach certain points can give a result
of infinity. Now we want to evaluate limits as we approach positive and negative infinity.
First, we need to state some useful facts that we will use often:
(
)
An intuitive proof would involve imagining that any number divided by a huge number (in
this case infinity) reduces to 0.
Also note that
(
)
(
)
(
)
Let’s look at some examples.
Example 4:
Evaluate
Solution:
(
)
( )
Example 5:
Evaluate
Solution: We know from the previous example that
So
Example 6:
Evaluate
Solution: In this case, we will first factor the numerator and the denominator.
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Computing Limits: Part II
(
)
(
)
Now we can use our properties for limits approaching infinity.
Notice that in both the numerator and denominator we factored out the variable with the
highest power. This is a common technique used to simplify functions which enables us
to apply of our defined properties.
Example 7:
Evaluate
Solution: To simplify the function, let’s factor out the highest power in both the
numerator and the denominator:
(
)
(
(
)
)
(
)
Notice that in this case, the power in the numerator is higher than the power in the
denominator. So the limit is:
(
)
(
(
)
)(
(
)
)
Example 8:
Evaluate
Solution:
(
)
(
)
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(
)
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Computing Limits: Part II
Notice that in this case, the power in the denominator is higher than the power in the
numerator. So the limit is:
(
( ) (
)
)
Example 9:
Evaluate
Solution: This may seem like a completely different problem, but in fact we go through
the same process by first factoring out the highest power.
(
)
(
(
)
)
(
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)
( )(
)
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Computing Limits: Part II
Exercises:
1. Evaluate the following limits:
a)
b)
√
c)
d)
2. Find the vertical asymptotes of the following functions:
a) ( )
b) f( )
c) ( )
d) ( )
( )
e) f( )
3. Evaluate the following limits:
a)
b)
c)
d)
e)
f)
g)
Solutions:
1. a) 12
b)
2. a) x= -5
b) x=1, x=-1
3. a)
b)
c)
d) 0
c) No asymptotes
c) 0
d)
Notice that in this case, we can directly plug in
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d) x=0
e) 2
e) x=-1
f) 0
g)
which gives us
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