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 evaluate the expression with the limit of , it returns the indeterminate form .

We can instead use L’Hospital’s Rule to evaluate, using the form:

Where,

Therefore,

If we rewrite the limit with L'Hospital's Rule,

To use L'hospital's rule, evaluate the limit of the numerator of the fraction and the denominator separately. If the result is , , or , take the derivative of the numerator and the denominator separately, and try to evaluate the limit again.

(L'hospital's rule)

(L'hospital's rule again)

(L'hospital's rule again)

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 .

If we evaluate the expression with the limit of x = 8, it returns the indeterminate form .

We can instead use L’Hospital’s Rule to evaluate, using the form:

Where,

Therefore,

If we rewrite the limit with 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.

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.

Evaluating the limit to begin with gets us , which is undefined. We can solve this problem using L'Hospital's rule. Taking the derivative of the numerator and denominator with respect to n, we get . The limit is still undefined. Another application of the rule gets us , which evaluated at is in fact .