# 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).

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.