L'Hospital's Rule

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.

The first thing we always have to do is to check that l'Hopital's rule is actually applicable when we want to use it.

So it is applicable here.

We take the derivative of the top and bottom, and get

and now we can safely plug in x=1 and get that the limit equals

.

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

.

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 again yields the indeterminate form

So we apply L'Hospital's rule again

Evaluating for yields

As such

and thus

When we evaluate the limit using normal methods, we arrive at the indeterminate form . When this occurs, to evaluate the limit, we must use L'Hopital's Rule, which states that

So, we must find the derivative of the top and bottom functions:

The derivatives were found using the following rule:

Now, rewrite the limit and evaluate it:

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:

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 .

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 .

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

When we evaluate the limit using normal methods, we get the indeterminate form . When this happens, we must use L'Hopital's Rule, which states that

Now, we must find the derivatives of the numerator and denominator:

The derivatives were found using the following rules:

, ,

Next, rewrite the limit and evaluate it: