L'Hospital's rule from Rudin's Principles. I am reading the proof of this theorem from Rudin and I have some questions.
1) Why denominator in (18) is not equal to $0$? I think that if $g(x)=g(y)$ then by Rolle's theorem $g'(\xi)=0$ for some point $\xi$. But for applying this theorem $g$ must be continuous at $[a,b]$. But our $g$ is differentiable in $(a,b)$ $\Rightarrow$ $g$ is continuous in $(a,b)$. 
2) Also why denominator in (19) i.e. $g(y)\neq 0$?
Can anyone help with these questions please?
 A: 1) If $g(x)=g(y)$ then there exists (by the mean value theorem) a $t$ between $x$ and $y$ with $g'(t)=0$, contradicting line one of the theorem.
2) Again, if $g(y)=0$, then there exists (by the mean value theorem) a $t$ between $a$ and $y$ such that $g'(t)=0$, contradicting line one of the theorem.
A: Intro
Rudin 5.13 (L'Hospital's rule). True, even for a book such as this, the change from (18) to (19) is... oh... well... This is how to do this bit. 
It starts at the words "Suppose (14) holds" at the bottom of page 109.

Warm up
We start by writing an expression, which is independent of the preceding text of the proof. We just write it first an then see what it means.
$$ \frac{f(y)-f(x)}{g(y)-g(x)} \tag{A} $$
Functions $f$ and $g$ were assumed differentiable on their domain $(a,b)$, therefore their quotient is also differentiable by Theorem 5.3 on page 103, and consequently it is also continuous by Theorem 5.2 on page 104, so the limit for (A) exists.
As we assumed $f(x) \to 0$ and $g(x) \to 0$ as $x \to a$ we may apply Theorem 4.4c on page 85 about the quotient of two limits, which in our case says that
$$ \lim_{x \to a} \frac{f(y)-f(x)}{g(y)-g(x)} = \frac{f(y)}{g(y)} \tag{B}$$ 
which equals some real number, we only need to find its whereabouts.
In order to do that we take (A), apply Generalized Mean Value Theorem (GMVT, Theorem 5.9 on page 107) to it to rewrite it as a quotient of differentials, and work with this quotient because it is much easier.
$$ \frac{f(y)-f(x)}{g(y)-g(x)} = \frac{f'(t)}{g'(t)} \tag{C}$$
$$ $$ 
Big picture
Up to now we have done the following. We have a relation between points
$$ \overbrace{a < \underbrace{x < t < y}_{GMVT} < c}^{\text{condition for (17)}} \tag{D}$$
for which the two key relations are (C) above and 
$$ \frac{f'(x)}{g'(x)} < r \tag{17} $$
What happens when we start pushing $x \to a$. Essentially we are constructing a sequence for values represented by the left side of (C). During this process for (A) we only change $f(x)$ and $g(x)$ that depend on $x$. 
The values of $f(y)$, $g(y)$ during these iterations remain the same, as well as $r$ because we fixed it when assuming the value of $A$ above, setting $r > A$.
Yet, each time we push $x \to a$ the value of $t$ in (C) changes! Each time we need to re-generate the new value of $t$ from each new segment $(x_n,y)$. Such points $t_n$ do not have to follow any pattern in this process, GMVT says there exist such points $t_n$ for each $(x_n, y)$.
On each iteration $x \to a$ we have (D) valid, so the sequence (C) may converge to any value strictly less than $r$. But $r$ may also be the limit point for this sequence, which the left side of (B) tries to approach. It is this fact that gives the sign of equality. For example
$$ s_n = r - \frac{1}{n}, \quad \quad \lim_{n \to +\infty} s_n = r $$
with $s_n$ strictly less than $r$ at each step. Therefore
$$ \frac{f(y)}{g(y)} \le r \tag{19} $$
$$ $$ 
By contradiction
There cannot be "strictly greater than sign" in (19), for in this case we are able to construct a sequence of (C) with (17) valid in each step, converging to some $r^{*} > r$. This $r^{*}$ will be the limit point for this sequence with infinitely many points in its neighborhood, which means infinitely many $t$'s for expression (C). Yet, for this sequence $t$ is confined by (17) in (D), a contradiction. There cannot be such $r^{*}$
$$ $$ 
Final points
Is there any effect on $t$, values and limit points of (C) as we let $x \to a$? Not really. GMVT puts no constraints on $t$ during this process. Points $t$ may "move chaotically" in $(x_n,y)$, get arbitrary close to $a$ or even $y$. 
What is important that $t$ always stays in  $(a,y)$. It cannot go below $a$ because in addition to (D) functions $f,g$ are not defined there.

