# L'Hopital's Rule

## Homework Statement

Evalutate the limit, as x approaches 0, log(coshx)/x2

L'Hopital's rule

## The Attempt at a Solution

I can get as far as (sinhx.cosh-1x)/2x by differentiating the top and bottom separately. I'm not sure how to do the next differentiation.

Thanks for any help.

Be careful about writing cosh-1 x; that means something quite different from what you mean.

So you have quite rightly differentiated top and bottom, leaving you with:

$$\frac{\frac{sinh x}{cosh x}}{2x}$$

Fix up that fraction so that it is in a more usual form (ie, move the cosh x to the denominator), and then apply L'Hopital's rule again.

Mark44
Mentor

## Homework Statement

Evalutate the limit, as x approaches 0, log(coshx)/x2

L'Hopital's rule

## The Attempt at a Solution

I can get as far as (sinhx.cosh-1x)/2x by differentiating the top and bottom separately. I'm not sure how to do the next differentiation.

Thanks for any help.

By cosh-1x, you really mean 1/cosh x, but this notation suggests the inverse cosh function. Don't write cosh-1x if you mean 1/cosh x.

Your numerator is sinh(x)/cosh(x) = tanh(x). Since the numerator approaches zero, and 2x approaches zero, use L'Hopital's Rule another time.

Here's what I've done. I've treated the numerator as being sinhx/coshx and differentiated it to get (1/cosh2x). I've treated denominator as being 2x and differentiated it to get 2. Therefore I end up with (1/cosh2x)/2 which gives 1/2 as x approaches 0. Is this the correct answer?

Yup.