Can the Stolz-Cesàro theorem be used to prove L'Hospital's rule, or viceversa? Stolz-Cesàro theorem is usually stated to be the discrete version of L'Hospital's rule. I was merely wondering whether one of these theorems could be used to prove the other (I couldn't find any proof that does this online).
 A: I found a proof of L 'Hopital's rule that makes use of Stolz-Cesaro for the case $g(x)\rightarrow \infty$.
I've noticed recently that in the hypothesis of L'Hopital's rule it is often assumed that $f$ and $g$ need be defined and differentiable at the point $p$ which is being approached in the limit (this is done in the proof I mentioned and in baby Rudin). Yet the theorem still holds even when $f$ and $g$ are not defined nor differentiable at $p$, so I will rewrite the proof with care so as to prove the theorem in the more general case.

Theorem: Let $f$ and $g$ be real functions differentiable in the open interval $I$ except possibly at the point $p\in I$. Assume that $g'(x)\neq 0$ for every $x\in I-\{p\}$ and that $$\lim_{x\rightarrow p}\frac{f'(x)}{g'(x)}$$ exists and equals $L$. If either
i) $\lim_{x\rightarrow p}g(x)=\pm \infty$, or
ii) $\lim{x\rightarrow p}f(x)=g(x)=0$,
then $\lim_{x\rightarrow p}f(x)/g(x)$ exists and equals $L$.

Proof:
i) Suppose $\lim_{x\rightarrow p}g(x)=\infty$. Consider all points $x\in I$ less than $p$, since $g'(x)\neq 0$ we have that $g'$ is either positive or negative, yet $\lim_{x\rightarrow p}g(x)=\infty$ implies it can only be positive. We may also assume $g(x)\neq 0$ by making $I$ smaller if necessary.
It suffices to show that for any monotone sequence $x_n\rightarrow p$  we get $f(x_n)/g(x_n)\rightarrow L$, so let $x_n$ be such a sequence. Regardless of whether it is approaching $p$ from the right or the left, we have that $g(x_{n+1})>g(x_n)$. Stolz-Cesaro gives
$$\liminf \frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}\leq \liminf \frac{f(x_n)}{g(x_n)}\leq \limsup \frac{f(x_n)}{g(x_n)}\leq \limsup \frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}$$
By Cauchy's Mean Value Theorem there exists $y_n$ between $x_n$ and $x_{n+1}$ such that $$\frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\frac{f'(y_n)}{g'(y_n)}$$
noting $\lim_{n\rightarrow \infty}f'(y_n)/g'(y_n)=L$ gives the desired result.
ii) Let $\epsilon >0$. Then there exists a neighborhood $U=(u,p)$ of $p$ such that $x\in U$ implies $$\Big| \frac{f'(x)}{g'(x)} -L\Big| <\epsilon$$
Let $y_n\rightarrow p$ be a monotone sequence in $U$. By Cauchy's Mean Value Theorem there is a sequence $z_n\in (x,y_n)$, so that $z_n\rightarrow p$, such that $$\frac{f(x)-f(y_n)}{g(x)-g(y_n)}=\frac{f'(z_n)}{g'(z_n)}\in (L-\epsilon,L+\epsilon)$$ therefore $\lim_{n\rightarrow \infty}\frac{f(x)-f(y_n)}{g(x)-g(y_n)}=\frac{f(x)}{g(x)}\in [L-\epsilon,L+\epsilon]$ for any $x\in U$, so $\lim_{x\rightarrow p^-}f(x)/g(x)=L$. The other direction is similar.
