2
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ 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)} $. $\endgroup$ – Abstraction May 24 '16 at 14:44
  • $\begingroup$ You are attempting to prove the converse to L'Hopital? $\endgroup$ – zhw. May 24 '16 at 17:00
  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh May 25 '16 at 9:59
  • $\begingroup$ @Abstraction: $h(x)$ is not continuous and there is no need for it to be continuous. $\endgroup$ – Paramanand Singh May 25 '16 at 10:00
  • $\begingroup$ 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 $x \to a^{+}$. $\endgroup$ – Paramanand Singh May 25 '16 at 10:01
3
$\begingroup$

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 $a=0$).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I didn't mean to be, I wrote the assumptions incorrectly. I'll make an edit. $\endgroup$ – Si.0788 May 24 '16 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.