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Proofs from previous slide

Souped Up Mean Value Theorem: If
$f(x)$ and $g(x)$ are continuous on a closed interval
$[a,b]$ and differentiable on the open interval $(a,b)$,
then there is a point $c$, between $a$ and $b$, where
$$\big(f(b)f(a)\big)g'(c) = \big(g(b)g(a)\big)f'(c). $$
(When $g(x)=x$, this is the same as the usual MVT.)
Proof: Consider the function $$
h(x) = \big(f(x)f(a)\big)\big(g(b)g(a)\big) 
\big(f(b)f(a)\big)\big(g(x)g(a)\big).$$ This is
continuous on $[a,b]$ and differentiable on $(a,b)$, with
$$h'(x) = f'(x)\big(g(b)g(a)\big)  g'(x) \big(f(b) 
f(a)\big).$$ Note that $h(a)=0=h(b)$. By Rolle's Theorem,
there a spot $c$ where $h'(c)=0$. But $h'(c)=
\big(f(b)f(a)\big)g'(c)  \big(g(b)g(a)\big)f'(c).$
Since this is zero, $\big(f(b)f(a)\big)g'(c) =
\big(g(b)g(a)\big)f'(c)$. $\qquad \hbox{QED}$

Proof of
Macho L'Hospital's Rule: By assumption, $f$ and $g$ are
differentiable to the right of $a$, and the limits of $f$
and $g$ as $x \to a^+$ are zero. Define $f(a)$ to be zero,
and likewise define $g(a)=0$. Since these values agree
with the limits, $f$ and $g$ are continuous on some
halfopen interval $[a,b)$ and differentiable on$(a,b)$.
For any $x \in (a,b)$, we have that $f$
and $g$ are differentiable on $(a,x)$ and continuous on
$[a,x]$. By the Souped up MVT, there is apoint $c$ between
$a$ and $x$ such that $f'(c) g(x) = f'(x) g(c)$. In other
words, $f'(c)/g'(c) = f(x)/g(x)$. Also, as $x$ approaches
$a$, $c$ also approaches $a$, since $c$ is somewhere
between $x$ and $a$. But then $$\lim_{x \to a^+}\,
\frac{f(x)}{g(x)} = \lim_{x \to a^+} \,\frac{f'(c)}{g'(c)}
= \lim_{c \to a^+}\,\frac{f'(c)}{g'(c)}.$$ That last
expression is the same as $\lim_{x \to a^+} f'(x)/g'(x)$.
$\qquad \hbox{QED}$
