TITLE: Proof of L'Hopital's rule
QUESTION [3 upvotes]: I'm attempting to prove L'Hopital's rule. My solution so far is the following:
Let $f,g:(a,b)\to\mathbb{R}$ be differentiable and continuous on $[a, b]$ with $f(a)=g(a)=0$ and $g(x)\neq0$ for $x\in(a, b)$. Suppose that $\lim_{x\to a^+}
\frac{f'(x)}{g'(x)}$ exists.
Let $x_n\in(a,b)$ be a sequence such that $\lim_{n\to\infty}x_n=a.$ 
Now by Cauchy's mean value theorem there exists $c_n\in(a,x_n)$ such that $$\frac{f'(c_n)}{g'(c_n)}=\frac{f(x_n)-f(a)}{g(x_n)-g(a)}=\frac{f(x_n)}{g(x_n)}$$
By the sandwich rule we also have that $\lim_{n\to\infty}c_n=a.$
I think I'm very nearly there but I can't see how to conclude it.

REPLY [3 votes]: It appears you are trying to prove the converse of L'Hopital: If $\lim f(x)/g(x) = L,$ then $\lim f'(x)/g'(x) = L.$ The converse is false. Take $f(x) = x^2\sin (1/x), g(x) = x$ to see this (here $a=0$).
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