Proving L'hospitals rule So I have one question concerning the proof of l'hospitals rule :
Assume that $f(x)\to \infty$ and $g(x) \to \infty $ and $\frac{f'(x)}{g'(x)}=L$ and $g'(x)$ is never zero.
Recall : We are trying to prove that $\frac{f(x)}{g(x)}=L$
Here In my book which is advanced analysis by Buck, The following is derived from the above :
"$g'(x)$ is never zero, we know that it must be positive and thus that g is strictly increasing; we may therefore assume $g(x)>0$ for all $x$"
I couldn't understand how we derived this, can't $g'(x)<0$? Why should it be positive $\forall x$?
Thanks
 A: By definition a function, $f$, is differentiable at $a$ if there is a function $\epsilon(x)$ so that ${\epsilon(x)\over x-a}\to 0 $ as $x\to a$ and

$$f(x) = f(a) + c(x-a) + \epsilon(x)$$

here we call the constant, $c$, the derivative of $f$ at $a$, and denote it by $f'(a)$.
With this definition it is fairly easy to see the result, let $\epsilon_f(x)$ and $\epsilon_g(x)$ be associated to $f,g$ respectively.
Then assume $f(a)=g(a)=0$ and of course both are assumed differentiable. Then we get

$$\lim_{x\to a} {f(x)\over g(x)}=\lim_{x\to a} {0+f'(a)(x-a) + \epsilon_f(x)\over 0+g'(a)(x-a)+\epsilon_g(x)}=\lim_{x\to a} {f'(a) + {\epsilon_f(x)\over x-a}\over g'(a) + {\epsilon_g(x)\over x-a}}={f'(a)\over g'(a)}$$

provided the denominator is not $0$, in which case if the numerator is $0$ we can use induction, and the limit does not exist otherwise. The case where the limit is infinite is the same, you simply make the numerator ${1\over g(x)}$ and the denominator ${1\over f(x)}$ and note that the ${0\over 0}$ case is equivalent.

Your book notes that derivatives are Darboux continuous, i.e. they have the intermediate value property, so that if $g'(x)>0$ anywhere and $g'(x)<0$ elsewhere it must be zero somewhere. So they reduce to the case $g$ is increasing.
