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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.

RamenChef
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    For $h(x) = {f'(x) \over g'(x)}$, $h$ is continuous on $(a,a+\varepsilon)$ (since $g'(a) \neq 0$). Also, let $\exists \lim_{x \to a^+}h(x)$ (when this is false, L'Hopital's rule doesn't work). You've proven that for your $(x_n)$ exists $(c_n) : c_n \to a, h(c_n) = {f(x_n) \over g(x_n)}$. By continuity $h(c_n) \to h(a)$, by definition of $(x_n)$ ${f(x_n) \over g(x_n)} \to \lim_{x \to a^+}{f(x) \over g(x)} $. – Abstraction May 24 '16 at 14:44
  • You are attempting to prove the converse to L'Hopital? – zhw. May 24 '16 at 17:00
  • Since $c_{n} \to a$ and $f'(x)/g'(x) \to L$ as $x \to a$ it follows that $f'(c_{n})/g'(c_{n}) \to L$ and therefore $f(x_{n})/g(x_{n}) \to L$ and since this happens for any arbitrary sequence $x_{n} \to a$ it follows that $f(x)/g(x) \to L$ as $x \to a$. This completes your proof. – Paramanand Singh May 25 '16 at 09:59
  • @Abstraction: $h(x)$ is not continuous and there is no need for it to be continuous. – Paramanand Singh May 25 '16 at 10:00
  • You don't need the condition that $g(x) \neq 0$ for all $x \in (a, b)$. This fact follows from the fact that $f'(x)/g'(x) \to L$ as $x \to a^{+}$. – Paramanand Singh May 25 '16 at 10:01

1 Answers1

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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 $a=0$).

zhw.
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