L'Hôpital's Rule

=Theorem= L'Hôpital's rule states that for any two functions, $$f(x), g(x) $$ where $$f'(x)$$ and $$g'(x)$$ exist in the neighborhood of c, $$\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}$$ exists, and $$f(c) = g(c) = 0 $$ or $$ f(c) = g(c) = \infty $$

$$ \lim_{x\rightarrow c} \frac{f(x)}{g(x)} = \lim_{x\rightarrow c} \frac{f'(x)}{g'(x)} $$

=Proof=

With the indeterminate form 0 over 0
This is the case where $$f(x) \to 0$$ and $$g(x) \to 0$$.

First, define (or redefine) $$f(c):=0$$ and $$g(c):=0$$. This does not change the limit, since the limit does not depend on the value at the point $$c$$ (by definition).

Take $$x$$ to be some point close to $$c$$. According to Cauchy's mean value theorem there is a point $$\xi$$ between $$x$$ and $$c$$ such that:



\frac{f'(\xi)}{g'(\xi)} = \frac{f(x) - f(c)}{g(x) - g(c)} $$

Since $$f(c) = g(c) = 0$$,


 * $$ \frac{f'(\xi)}{g'(\xi)} = \frac{f(x)}{g(x)}

$$

If $$x\to c$$, then also $$\xi \to c$$ and



\lim_{x\to c}\frac{f'(x)}{g'(x)} = \lim_{\xi\to c}\frac{f'(\xi)}{g'(\xi)} = \lim_{x\to c}\frac{f(x)}{g(x)}, $$ as required.