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

.

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.

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 .

Directly evaluating for yields the indeterminate form

we are able to apply L'Hospital's rule which states that if the limit is in indeterminate form when evaluated, then

As such the limit in the problem becomes

Evaluating for yields

As such

and thus

Directly evaluating for yields the indeterminate form

we are able to apply L'Hospital's rule which states that if the limit is in indeterminate form when evaluated, then

As such the limit in the problem becomes

Evaluating for yields

As such

and thus

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 .