I have a little problem understanding a point of Rudin's proof on L'hopital's Rule. Here is the statement of the L'hospital's Rule:
L'hospital's Rule Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)\neq 0$ for all $x\in (a,b)$, where $-\infty \leq a<b\leq \infty$. Suppose $\frac{f'(x)}{g'(x)}\rightarrow A$ as $x\rightarrow a$. If $f(x)\rightarrow 0$ and $g(x)\rightarrow 0$ as $x\rightarrow a$, then $\frac{f(x)}{g(x)} \rightarrow A$ as $x\rightarrow a$.
Rudin uses the generalized mean value theorem (GMVT):
GMVT If f and g are continuous real functions on [a, b] which are differentiable in (a, b), then there is a point $x\in (a,b)$ at which $[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)$.
Here is Rudin's proof on L'hospital's Rule:
Proof: We first consider the case in which $-\infty \leq A<\infty$. Choose a real number $q$ such that $A<q$, and then choose $r$ such that $A<r<q$. Then, there is a point $c\in (a,b)$ such that $a<x<c$ implies $\frac{f'(x)}{g'(x)}<r$. If $a<x<y<c$, then GMVT shows that there is a point $t\in (x, y)$ such that $\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(t)}{g'(t)}<r$.
For the last line of the above paragraph, I am a little confused on how can we make sure that $g(x)\neq g(y)$? Even though $g'(x)\neq 0$ for all $x\in (a,b)$, we still may have $g(x)=g(y)$ for some $x$ and $y$ in $(a,c)$. If $g(x)=g(y)$, we have a fraction over $0$ which can go to infinity, and thus cannot be less than $r\in R$.
Best Answer
By Rolle's Theorem, if $g(x)=g(y)$ for some $x <y$, then, there exists some $c \in (x,y)$ such that $g'(c)=0$.