If you plug in the limit value, the function turns into .  Therefore, we are allowed to use l'Hospital's Rule.  We start by taking the derivative of both the numerator and the denominator until when you plug in the value of the limit, you do not get something in the form .  Luckily, in this case, we only need to take the derivative once.  By taking the derivatives separately, we get a new limit: 

.

We rewrite the limit as

Substituting  yields the indeterminate form 

L'Hospital's Rule states that when the limit is in indeterminate form, the limit becomes

For  and  we solve the limit

and substituting  we find that 

As such

Substituting  yields the indeterminate form 

L'Hospital's Rule states that when the limit is in indeterminate form, the limit becomes

For  and  we solve the limit

and substituting  we find that 

As such

Simply substituting  in the given limit will not work:

Because direct substitution yields an indeterminate result, we must apply L'Hospital's rule to the limit:

if and only if  and both  and  exist at .

Here,

 and .

Hence, 

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:

, , 

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:

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.

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 .

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.