Stolz-Cesaro Theorem, 0/0 Case
How to prove the $0/0$ case of Stolz-Cesaro Theorem? In other words:
Given that $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = 0$$ with $(b_n)_{n=1}^{\infty}$ strictly monotone, and that $$\lim_{n \to \infty} \frac{a_{n+1} - a_{n}}{b_{n+1} - b_{n}} = L$$ prove that $\lim_{n \to \infty} \frac{a_n}{b_n} = L$.
Without loss of generality, suppose ${b_n}$ monotonically decreases. Thus $b_n\gt0$ and $b_n-b_{n+1} \gt 0$.
If $$\lim_{n\to \infty}\frac{a_n-a_{n+1}}{b_n-b_{n+1}}=L,$$
then for any given $\epsilon \gt 0$, there exists a natural number $N$ such that
$$(L-\epsilon)(b_n-b_{n+1}) \lt a_n-a_{n+1} \lt (L+\epsilon)(b_n-b_{n+1})$$
for all $n \gt N$.
Let $k \gt n$ be a natural number. Summing the inequalities above, we get:
$$(L-\epsilon)\sum_{i=n}^{k}(b_i-b_{i+1}) \lt \sum_{i=n}^{k}(a_i-a_{i+1})
\lt (L+\epsilon)\sum_{i=n}^{k}(b_i-b_{i+1}),$$
i.e.,$$(L-\epsilon)(b_n-b_{k+1}) \lt a_n-a_{k+1} \lt (L+\epsilon)(b_n-b_{k+1}).$$
In this inequality, fix $n$ and let $k \to \infty$, we get
$$(L-\epsilon)b_n \le a_n \le (L+\epsilon)b_n,$$
that is, $$L-\epsilon \le \frac{a_n}{b_n} \le L+\epsilon.$$
This relation holds for all $n \gt N$, so finally $$\lim_{n \to \infty}\frac{a_n}{b_n}=L.$$