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Example Questions
Example Question #36 : L'hospital's Rule
Evaluate the following limit:
When evaluating the limit using normal methods (substitution), we receive the indeterminate form . When we receive the indeterminate form, we must use L'Hopital's Rule to evaluate the limit. The rule states that
Using the formula above for our limit, we get
The derivatives were found using the following rules:
, ,
Example Question #32 : L'hospital's Rule
Evaluate the limit:
When evaluating the limit using normal methods (substitution), we get the indeterminate form . When this happens, to evaluate the limit we use L'Hopital's Rule, which states that
Using the above formula for our limit, we get
The derivatives were found using the following rule:
Example Question #31 : L'hospital's Rule
Evaluate the following limit, if possible:
The limit does not exist.
To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get , which is undefined. We now use L'Hopital's rule which says that if and are differentiable and
,
then
.
We are evaluating the limit
.
In this case we have
and
.
We differentiate both functions and find
and
By L'Hopital's rule
.
When we plug the limit value of 2 into this expression we get 9/3, which simplifies to 3.
Example Question #39 : L'hospital's Rule
Evaluate the following limit, if possible:
.
The limit does not exist.
If we plugged in the limit value, , directly we would get the indeterminate value . We now use L'Hopital's rule which says that if and are differentiable and
,
then
.
The limit we wish to evaluate is
,
so in this case
and
.
We calculate the derivatives of both of these functions and find that
and
.
Thus
.
When we plug the limit value, , into this expression we get , which is .
Example Question #32 : L'hospital's Rule
Evaluate the following limit, if possible:
.
The limit does not exist
The limit does not exist
We will show that the limit does not exist by showing that the limits from the left and right are different.
We will start with the limit from the right. Using the product rule we rewrite the limit
.
We know that
and
so
.
We calculate the limit from the left in the same way and find
.
Thus the two-sided limit does not exist.
Example Question #41 : L'hospital's Rule
Evaluate the following limit, if possible:
.
The limit does not exist
To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get , which is undefined. We now use L'Hopital's rule which says that if and are differentiable and
,
then
.
In this case we are calculating
so
and
.
We calculate the derivatives and find that
and
.
Thus
.
Example Question #42 : L'hospital's Rule
Evaluate the following limit, if possible:
The limit does not exist
The limit does not exist
The limit does not exist. When we first plug in the limit value, 0, we get 0/0 and indeterminate form. We may try to use L'Hopital's rule. However, we quickly discover that
is not differentiable at x=0, so we cannot use L'Hopital's rule. We then consider the limits from the left and right. We determine that
and
.
Thus the two sided limit does not exist.
Example Question #43 : L'hospital's Rule
Evaluate the following limit, if it exists:
To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get , which is undefined. We now use L'Hopital's rule which says that if and are differentiable and
,
then
.
We are trying to evaluate the limit
,
so we have
and
.
We differentiate both of these functions and find
and
.
Using L'Hopital's rule we see
.
Now when we plug the limit value into the expression we get , so
.
Example Question #44 : L'hospital's Rule
Evaluate the following limit, if possible:
.
The limit does not exist.
If we plugged in the limit value, , directly we would get the indeterminate value . We now use L'Hopital's rule which says that if and are differentiable and
,
then
.
The limit we wish to evaluate is
,
so we have
and
.
We evaluate the derivatives of these two functions and find
and
.
Thus, using L'Hopital's we find
.
However, when we plug the limit value into this second expression we still end up with the indeterminate value . So we use L'Hopital's again.
and
.
Using L'Hopital's rule again we see
.
Plugging in the limit value now we see that the limit evaluates to
Example Question #45 : L'hospital's Rule
Evaluate the following limit, if possible:
The limit does not exist
To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get , which is undefined. We now use L'Hopital's rule which says that if and are differentiable and
,
then
.
We are trying to evaluate the limit
so we have
and
.
We calculate the derivatives of these functions and get
and
.
Using L'Hopital's rule we find
.
Now when we plug in the limit value we get , so the limit evaluates to 0.
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