Why does 'Hopital not work here?

Question

Can you find the this limit using l'Hopital? If not then why not? I am confused on whether it works here and how I would know not to use it.

lim x->0 (1-cos4x) / (1-cos6x)

Answer 1

Hi R.R.

L'Hopital DOES work here, it just must be repeated a few times.

because limit as x approaches 0 of (1-cos4x)/(1-cos6x) = 0/0, L'Hopital's Rule will work

take derivative of top, and derivative of bottom separately.

first time the result is lim as x approaches 0 of (4 sin4x)/(6 sin6x) = 0/0 so LH rule will still work.

do it again.

take derivative of top and derivative of bottom separately

second time the result is limit as x approaches 0 of (16 cos4x)/(36 cos 6x) = 16/36 = 4/9

I graphed this on desmos to be sure, but I don't think I can attach it here. Here is the link:

https://www.desmos.com/calculator/op47kgc93d

Answer 2

0/0 in a limit can be resolvable using l'hopital once or twice or until the resolution is found.

here first application is lim x->0 4sin4x/6sin6x which is 0/0

second application is lim x->0 16cos4x/36cos6x which is 16/36

If you graph (1-cos4x)/(1-cos6x) you'll see the value at x=0.

It looks interesting at x=pi/2, note that there that the function numerator is zero and denominator 2.