Clear proof of L'Hopital rule for $\infty/\infty$ form I am trying to understand the proof for L'Hopital rule for the $\infty/\infty$ form given here. It is really clear for most part except the last three steps, where it whizzes rapidly out of my grasp. How did they get that inequality mentioned there? Can someone please provide a simple explanation for those steps? Alternatively, is there another simpler proof for this?
Thanks.
 A: You first proved, using Cauchy's Mean Value Theorem, that
$$\color{blue}{\frac{f(x)}{g(x)}=\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}\cdot\frac{f'(\xi_x)}{g'(\xi_x)}}$$
and also
$$\color{red}{\left|\frac{f'(\xi)}{g'(\xi)}-m\right|<\epsilon}\,,\,\,\text{for}\;\;0<|\xi-a|<\delta$$
and also, that
$$\lim_{x\to a}\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}=1\implies\color{green}{\left|\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}-1\right|<\frac\epsilon{|m|+\epsilon}}\;,\;\;\text{for}\;\;0<|x-a|<\delta''<\delta$$
and from here
$$\left|\frac{f(x)}{g(x)}-m\right|\stackrel{\color{blue}{(*)}}=\left|\color{red}{\left(\frac{f'(\xi_x)}{g'(\xi_x)}-m\right)}+\frac{f'(\xi_x)}{g'(\xi_x)}\color{green}{\left(\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}-1\right)}\right|$$$${}$$
where $\color{blue}{(*)}\;$ is the first, blue, equality above (just open up parentheses and check!). You now have only to substitute the red and green inequalities
A: Here is another proof for the $\infty/ \infty$ case as $x \to a+,$ which I find easier to follow.
Note that we only need $g(x) \to \infty$.
If 
$$L = \lim_{x \to a+}\frac{f'(x)}{g'(x)},$$
then, for any $\epsilon > 0$ there exists $\delta_1 > 0 ,$ such that 
$$L - \epsilon< \frac{f(x)- f(y)}{g(x) - g(y)} = \frac{f'(\xi)}{g'(\xi)} < L + \epsilon,$$
for $a < x < \xi <y < a + \delta_1,$ where the appearance of $\xi \in (x,y)$ follows from the MVT.
Hence,
$$L - \epsilon < \frac{\frac{f(x)}{g(x)}- \frac{f(y)}{g(x)} }{1- \frac{g(y)}{g(x)} } < L + \epsilon, $$
and
$$ L - \epsilon  - \underbrace{\frac{(L - \epsilon)g(y)-f(y)}{g(x)}}_{C(x)}<  \frac{f(x)}{g(x)} < L + \epsilon + \underbrace{\frac{(L + \epsilon)g(y)+f(y)}{g(x)}}_{D(x)}.  $$
Since $g(x) \to \infty$ there exists $\delta_2$ such that if $a < x < a + \delta_2$, with $y$ fixed,  we have $-\epsilon < C(x), \,\, D(x) < \epsilon.$
Hence, if $a < x < a+ \min(\delta_1,\delta_2),$ then we have
$$L - 2\epsilon < \frac{f(x)}{g(x)} < L + 2\epsilon.$$
A: This is taken from my blog post.

There is another version of L'Hospital's Rule which is not so widely known and we state (and prove) it below:
If $f(x), g(x)$ are differentiable in a certain neighborhood of $a$ (but not necessarily at $a$), $\lim_{x \to a}\dfrac{1}{g(x)} = 0$ (equivalently $|g(x)| \to \infty$ as $x \to a$) and $\lim_{x \to a}\dfrac{f'(x)}{g'(x)} = L$ then $\lim_{x \to a}\dfrac{f(x)}{g(x)} = L$.
Thus in order to apply this version of L'Hospital's Rule we need to check that $|g(x)| \to \infty$ as $x \to a$. No check apart from differentiability is needed for the function $f(x)$. We prove this rule using the $\epsilon, \delta$ definition of limit. Since $f'(x)/g'(x) \to L$ as $x \to a$, it follows that $f'(x)/g'(x)$ is bounded in a certain deleted neighborhood of $a$. Therefore there is a number $A > 0$ and a number $\delta_{1} > 0$ such that $$\left|\frac{f'(x)}{g'(x)}\right| < A$$ for all $x$ with $0 < |x - a| < \delta_{1}$. Let $\epsilon > 0$ be arbitrary. Then we know that there is a $\delta_{2} > 0$ such that $$\left|\frac{f'(x)}{g'(x)} - L\right| < \frac{\epsilon}{3}$$ for all $x$ with $0 < |x - a| < \delta_{2}$.
Let's consider the ratio $$\frac{f(x) - f(y)}{g(x) - g(y)}$$ where both $x, y$ are distinct points lying in deleted neighborhood $(a - \delta_{3}, a + \delta_{3}) - \{a\}$ of $a$ and $\delta_{3} = \min(\delta_{1}, \delta_{2})$. We can express this ratio as $$\frac{f(x) - f(y)}{g(x) - g(y)} = \dfrac{\dfrac{f(x)}{g(x)} - \dfrac{f(y)}{g(x)}}{1 - \dfrac{g(y)}{g(x)}}$$ and from this equation we obtain $$\frac{f(x)}{g(x)} = \frac{f(x) - f(y)}{g(x) - g(y)}\left(1 - \frac{g(y)}{g(x)}\right) + \frac{f(y)}{g(x)} = \frac{f'(c)}{g'(c)}\left(1 - \frac{g(y)}{g(x)}\right) + \frac{f(y)}{g(x)}$$ where $c$ is some number between $x$ and $y$. Let $y$ have a fixed value in the deleted neighborhood $(a - \delta_{3}, a + \delta_{3}) - \{a\}$ of $a$. Then we know that $g(y)/g(x) \to 0, f(y)/g(x) \to 0$ as $x \to a$. Hence there are positive numbers $\delta_{4}, \delta_{5}$ such that $$\left|\frac{g(y)}{g(x)}\right| < \frac{\epsilon}{3A}$$ for all $x$ with $0 < |x - a| < \delta_{4}$ and $$\left|\frac{f(y)}{g(x)}\right| < \frac{\epsilon}{3}$$ for all $x$ with $0 < |x - a| < \delta_{5}$. Let $\delta = \min(\delta_{3}, \delta_{4}, \delta_{5})$ and let $0 < |x - a| < \delta$ then we have
\begin{align}
\left|\frac{f(x)}{g(x)} - L\right| &= \left|\frac{f'(c)}{g'(c)}\left(1 - \frac{g(y)}{g(x)}\right) + \frac{f(y)}{g(x)} - L\right|\notag\\
&= \left|\frac{f'(c)}{g'(c)} - L - \frac{f'(c)}{g'(c)}\cdot\frac{g(y)}{g(x)} + \frac{f(y)}{g(x)}\right|\notag\\
&\leq \left|\frac{f'(c)}{g'(c)} - L\right| + \left|\frac{f'(c)}{g'(c)}\right|\left|\frac{g(y)}{g(x)}\right| + \left|\frac{f(y)}{g(x)}\right|\notag\\
&< \frac{\epsilon}{3} + A\cdot\frac{\epsilon}{3A} + \frac{\epsilon}{3}\notag\\
&= \epsilon\notag
\end{align}
It now follows that $f(x)/g(x) \to L$ as $x \to a$ and the proof of the second version of L'Hospital's Rule is complete. Tt is easy to prove in similar manner that if $f'(x)/g'(x)$ tends to $\infty$ (or to $-\infty$) then so does $f(x)/g(x)$.
