Stolz-Cesàro $0/0$ case: is $\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$? The general form of Stolz-Cesaro $\infty/\infty$ case states that any two real two sequences $a_n$ and $b_n$, with the latter being monotone and unbounded, satisfy
$$\liminf\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\le\liminf\frac{a_n}{b_n}\le\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$Does the same hold for the $0/0$ case? That is, is it true that if $\lim a_n=\lim b_n=0$ and $b_n$ is strictly monotone, then $$\liminf\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\le\liminf\frac{a_n}{b_n}\le\limsup \frac{a_n}{b_n}\le \limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$?
EDIT: Here's my attempt, please any feedback is appreciated.
I tried with the $\limsup$, assuming $0<b_{n+1}<b_n$ for all $n$. Suppose $\alpha>\limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$. 
Then there exist infinitely many $N$ such that for all $k\ge0$, $$\alpha>\frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}.$$
Since $b_{n+1}<b_n$, we have for $k\ge0$ that $\alpha(b_{N+k}-b_{n+K-1})<a_{N+k}-a_{N+k-1}$.
Thus for any $m\ge0$, \begin{align} \alpha\sum_{k=0}^m(b_{N+k}-b_{N+k-1}) &< \sum_{k=0}^m(a_{N+k}-a_{N+k-1}) \\ \alpha(b_{N+m}-b_{N-1})&<a_{N+m}-a_{n-1}\end{align}and taking $m\to\infty$, \begin{align} -\alpha b_{N-1}&<-a_{N-1}  \\ \alpha&>\frac{a_{N-1}}{b_{N-1}}.\end{align}Taking finally $N\to\infty$, we must have $\alpha\ge\limsup_{n\to\infty}\frac{a_n}{b_n}$. Thus we can conclude $$\limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\ge\limsup_{n\to\infty}\frac{a_n}{b_n}.$$
 A: This is a nice (and correct) argument. As such, I only have a couple of minor remarks regarding how the proof is presented (which you can feel free to ignore).


*

*The assumption $0<b_{n+1}<b_n$ is an additional assumption (or rather $(b_n)_n$ being a positive sequence is an assumption and the inequality then is a consequence of monotony). I think it would be best to write "WLOG" (and maybe argue why that's the case) to make clear what you are doing.

*It might be clearer to write "Let $\alpha>...$ be arbitrary" to emphasize that you are talking about any such $\alpha$.

*In particular, when you say "Suppose $\alpha>\limsup_{n\rightarrow}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$", you assume the $\limsup$ takes a real value. This can't be the case when it is infinity, so you should probably add a short note saying that you  assume it is real, because the inequality is trivial when the $\limsup$ is infinite.

*The formulation "There exist infinitely many $N$ such that for all $k\ge0$, $\alpha>\frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$" seems weirdly redundant, as once you have once such $N$, it holds for any $N^\prime\ge N$, so it would be clearer to say something like $\alpha>\frac{a_{N^\prime+k}-a_{N^\prime+k-1}}{b_{N^\prime+k}-b_{N^\prime+k-1}}$ for all $N^\prime\ge N$ and $k\ge0$ (this would also make taking the limit as $N^\prime\rightarrow\infty$ clearer).

*In the next line it becomes apparent that choosing the symbol $\alpha$ when you also have a sequence $a_n$ was probably suboptimal in terms of being easy to read.

*Taking the limit of convergent sequences preserves order, but it doesnt strictly does so (you can have $a_n<b_n$, but $\lim a_n=\lim b_n$), so the inequalities $-\alpha b_{N-1}<-a_{N-1}$ and $\alpha>\frac{a_{N-1}}{b_{N-1}}$ shouldn't be strict (this is the only mathematical mistake, but it doesn't affect the result).

*Lastly, you have proven $\limsup_{n\rightarrow\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\ge\limsup_{n\rightarrow\infty}\frac{a_n}{b_n}$, but the part
$$\limsup_{n\rightarrow\infty}\frac{a_n}{b_n}\ge\liminf_{n\rightarrow\infty}\frac{a_n}{b_n}\ge\liminf_{n\rightarrow\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
is still missing. The middle inequality is trivial, so you might as well ignore it (or remark that it is indeed trivial), but you should at least note that the $\liminf$ inequality follows directly from the $\limsup$ inequality (and maybe how so).

