Prove the following: Let $\displaystyle a<b $ be real numbers, let $\displaystyle f: [a,b] \to \bold{R} $ and $\displaystyle g: [a,b] \to \bold{R} $ be functions which are differentiable on $\displaystyle [a,b] $. Suppose that $\displaystyle f(a) = g(a) = 0 $, that $\displaystyle g' $ is nonzero on $\displaystyle [a,b] $, and $\displaystyle \lim_{x \to a: x \in (a,b]} \frac{f'(x)}{g'(x)} $ exists and equals $\displaystyle L $. Then $\displaystyle g(x) \neq 0 $ for all $\displaystyle x \in (a,b] $, and $\displaystyle \lim_{x \to a: x \in (a,b]} \frac{f(x)}{g(x)} $ exists and equals $\displaystyle L $.

Suppose that $\displaystyle g(x) = 0 $ for some $\displaystyle x \in (a,b] $. Since $\displaystyle g(a) = 0 $, then by Rolle's theorem we can find a $\displaystyle y \in (a,b] $ such that $\displaystyle g'(y) = 0 $. But this contradicts the fact that $\displaystyle g' $ is nonzero on $\displaystyle [a,b] $. Hence $\displaystyle g(x) \neq 0 $ for all $\displaystyle x \in (a,b] $.

Now to show that $\displaystyle \lim_{x \to a: x \in (a,b]} \frac{f'(x)}{g'(x)} $ exists and equals $\displaystyle L $, we can show that $\displaystyle \lim_{n \to \infty} \frac{f(x_{n})}{g(x_{n})} = L $ for any sequence $\displaystyle (x_{n})_{n=1}^{\infty} $ which converges to $\displaystyle x $. How would I show this? I am guessing that I have to create a new function, and somehow manipulate it so that $\displaystyle \frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)} $. And this is equivalent to showing that $\displaystyle \frac{f(x_{n})}{g(x_{n})} = \frac{f'(x_{n})}{g'(x_{n})} $.