Question

Let f and g be functions that are differentiable for all real numbers, with g(x)≠0 for x≠0. If lim x->0 f(x)=lim x->0 g(x)=0 and lim x->0 f'(x)/g'(x) exists, then lim x->0 f(x)/g(x)=?

Answer 1

Please help!!

Answer 2

This is an application of L'hopitals rule. Do you remember what it says?

Answer 3

Yes but I'm still finding it hard to understand the question.

Answer 4

So we know that \(\lim_{x\to0}f(x)=g(x)=0\). So \[\lim_{x\to0}\frac{f(x)}{g(x)}\]Is of the form \(0/0\). So we can apply L'hopitals rule to say that\[\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}.\]Do you see what the answer is now?

Answer 5

It is C then?

Answer 6

That's exactly it. Good job.

Answer 7

Alright that was simpler than I thought. Thank you very much! @KingGeorge

Answer 8

You're welcome.