All Calculus 2 Resources
Example Questions
Example Question #91 : Gre Subject Test: Math
Find the limit:
In order to determine the limit, substitute to determine whether the expression is indeterminate.
Use the L'Hopital's rule to simplify. Take the derivative of the numerator and denominator separately, and reapply the limit.
Substitute
Example Question #31 : New Concepts
Evaluate:
The limit does not exist.
By substitution, the limit will yield an indeterminate form . L'Hopital can be used in this scenario.
Take the derivative of the numerator and denominator separately, and then reapply the limit.
The answer is .
Example Question #202 : Derivatives
Evaluate the following limit:
When you try to solve the limit using normal methods, you find that the limit approaches zero in the numerator and denominator, resulting in an indeterminate form "0/0".
In order to evaluate the limit, we must use L'Hopital's Rule, which states that:
when an indeterminate form occurs when evaluting the limit.
Next, simply find f'(x) and g'(x) for this limit:
The derivatives were found using the following rules:
,
Next, using L'Hopital's Rule, evaluate the limit using f'(x) and g'(x):
Example Question #11 : L'hospital's Rule
Determine the limit of:
Undefined
Rewrite the expression.
By substitutition, we will get the indeterminate form .
The L'Hopital's rule can be used to solve for the limit. Write the L'Hopital's rule.
Apply this rule twice.
Example Question #32 : New Concepts
Evaluate the following limit:
When evaluating the limit using normal methods (substitution), you find that the indeterminate form of is reached. To evaluate the limit, we can use L'Hopital's Rule, which states that:
So, we find the derivative of the numerator and denominator, which is
and , respectively.
The derivatives were found using the following rules:
,
When we evaluate this new limit, we find that
.
Example Question #12 : L'hospital's Rule
Evaluate the limit:
When we evaluate the limit using normal methods (substitution), we get an indeterminate form . To evaluate the limit, we must use L'Hopital's Rule, which states that:
Now we compute the derivative of the numerator and denominator of the original function:
,
The derivatives were found using the following rules:
,
Now, evaluate the limit:
Example Question #33 : New Concepts
Evaluate the following limit:
The limit does not exist
When we evaluate the limit using normal methods, we get , an indeterminate form. So, to evaluate the limit, we must use L'Hopital's Rule, which states that when we receive the indeterminate form of the type above (or ):
So, we must find the derivative of the top and bottom functions:
,
The derivatives were found using the following rules:
, ,
Now, rewrite the limit and evaluate:
Example Question #34 : New Concepts
Evaluate the limit:
The limit does not exist
When evaluating the limit using normal methods, we get the indeterminate form . When this occurs, or when occurs, we must use L'Hopital's Rule to evaluate the limit. The rules states
So, we must find the derivative of the numerator and denominator:
The derivatives were found using the following rules:
, ,
Now, using the above formula, evaluate the limit:
Example Question #92 : Calculus
Find the limit if it exists.
Hint: Apply L'Hospital's Rule.
Through direct substitution, we see that the limit becomes
which is in indeterminate form.
As such we can use l'Hospital's Rule, which states that if the limit
is in indeterminate form, then the limit is equivalent to
Taking the derivatives we use the power rule which states
Using the power rule the limit becomes
As such the limit exists and is
Example Question #22 : Limits
Find the limit if it exists.
Hint: Apply L'Hospital's Rule.
Through direct substitution, we see that the limit becomes
which is in indeterminate form.
As such we can use l'Hospital's Rule, which states that if the limit
is in indeterminate form, then the limit is equivalent to
Taking the derivatives we use the trigonometric rule which states
where is a constant.
Using l'Hospital's Rule we obtain
And through direct substitution we find
As such the limit exists and is
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