# L'Hopital's Rule

• Nov 11th 2009, 08:27 PM
jzellt
L'Hopital's Rule
Can anyone show a simple proof of L'Hopital's Rule?

lim as x->0 of f(x)/g(x) = f'(0)/g'(0)

Thanks...
• Nov 11th 2009, 08:53 PM
redsoxfan325
Quote:

Originally Posted by jzellt
Can anyone show a simple proof of L'Hopital's Rule?

lim as x->0 of f(x)/g(x) = f'(0)/g'(0)

Thanks...

The proof of the general case is a bit ugly (at least the one I know is). Here's a simple proof of the $\displaystyle \frac{0}{0}$ case of the rule:

Theorem: If $\displaystyle f(x_0)=0=g(x_0)$, $\displaystyle f'(x_0)$ and $\displaystyle g'(x_0)$ exist and do not both equal $\displaystyle 0$ or $\displaystyle \infty$, then $\displaystyle \lim_{x\to x_0}\frac{f(x)}{g(x)}=\frac{f'(x_0)}{g'(x_0)}$.

Proof: $\displaystyle \lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{g(x)-g(x_0)}=\lim_{x\to x_0}\frac{\displaystyle\frac{f(x)-f(x_0)}{x-x_0}}{\displaystyle\frac{g(x)-g(x_0)}{x-x_0}}=$ $\displaystyle \frac{\displaystyle\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}}{\displaystyle\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}}=\frac{f'(x_0)}{g'(x_0)}$

The $\displaystyle \frac{\infty}{\infty}$ case is trickier.
• Nov 11th 2009, 10:39 PM
Drexel28
redsoxfan325's proof is quite a common reasoning, but for a rigorous proof you need a bit of machinery. Namely, what really is a derivative? etc.

@redsoxfan325- I of course in no way am implying that you don't know that (I have seen your posts and you obviously do) but am merely making an observation for the OP.