Substitution is invalid.  In order to solve , rewrite this as an equation.

Take the natural log of both sides to bring down the exponent.

Since  is in indeterminate form, , use the L'Hopital Rule.

L'Hopital Rule is as follows:

This indicates that the right hand side of the equation is zero.

Use the term  to eliminate the natural log.

L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get

. 

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

.

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

 and . 

So we can simplify the function by remembering that any number divided by infinity gives you zero.

L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get 

. 

Since the first set of derivatives eliminates an x term, we can plug in zero for the x term that remains. We do this because the limit approaches zero.

This gives us

.

L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get 

. 

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

.

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

. 

To calculate the limit, often times we can just plug in the limit value into the expression. However, in this case if we were to do that we get , which is undefined.

What we can do to fix this is use L'Hopital's rule, which says

.

So, L'Hopital's rule allows us to take the derivative of both the top and the bottom and still obtain the same limit.

.

Plug in  to get an answer of .

If we plugged in the integration limit to the expression in the problem we would get , which is undefined. Here we use L'Hopital's rule, which is shown below.

This gives us, 

.

However, even with this simplified limit, we still get . So what do we do? We do L'Hopital's again!

.

Now if we plug in infitinity, we get 0. 

If we plugged in  directly, we would get an indeterminate value of .

We can use L'Hopital's rule to fix this. We take the derivate of the top and bottom and reevaluate the same limit.

.

We still can't evaluate the limit of the new expression, so we do it one more time.

Subbing in zero into  will give you , so we can try to use L'hopital's Rule to solve.

First, let's find the derivative of the numerator. 

 is in the form , which has the derivative , so its derivative is . 

 is in the form , which has the derivative , so its derivative is .

The derivative of  is  so the derivative of the numerator is .

In the denominator, the derivative of  is , and the derivative of  is . Thus, the entire denominator's derivative is .

Now we take the 

, which gives us . 

If we plug in 0 into the limit we get , which is indeterminate.

We can use L'Hopital's rule to fix this. We can take the derivative of the top and bottom and reevaluate the limit.

.

Now if we plug in 0, we get 0, so that is our final limit.

If we try to directly plug in the limit value into the function, we get

Because the limit is of the form , we can apply L'Hopital's rule to "simplify" the limit to

.

Now if we directly plug in 0 again, we get

.