stolz定理：若

$(1)y_{n+1}>y_n\qquad(n=1,2,\cdots);\\(2)\lim\limits_{n\to\infty}y_n=+\infty;$

$(3)\lim\limits_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}$存在，

则$\lim\limits_{n\to\infty}\frac{x_n}{y_n}=\lim\limits_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}.$

证：设$\lim\limits_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=a,$

由此，并注意到$\lim\limits_{n\to\infty}y_n=+\infty$,知对于任意给定的$\varepsilon>0$,存在自然数$N$,使得当$n>N$时，恒有

$$|\frac{x_{n+1}-x_n}{y_{n+1}-y_n}-a|<\varepsilon，$$且$y_n>0$.

即   $a-\varepsilon<\frac{x_{n+1}-x_n}{y_{n+1}-y_n}<a+\varepsilon\qquad (n=N+1,N+2,\cdots).$

又因为$y_{n+1}>y_n,$所以有

       $(a-\varepsilon)(y_{N+2}-y_{N+1})<x_{N+2}-x_{N+1}<(a+\varepsilon)(y_{N+2}-y_{N+1})\\(a-\varepsilon)(y_{N+3}-y_{N+2})<x_{N+3}-x_{N+2}<(a+\varepsilon)(y_{N+3}-y_{N+2})$

              $\cdots$

       $(a-\varepsilon)(y_{n+1}-y_n)<x_{n+1}-x_n<(a+\varepsilon)(y_{n+1}-y_n).$

从而$(a-\varepsilon)(y_{n+1}-y_{N+1})<x_{n+1}-x_{N+1}<(a+\varepsilon)(y_{n+1}-y_{N+1}),$

即$$(a-\varepsilon)(1-\frac{y_{n+1}}{y_{n+1}})+\frac{x_{N+1}}{y_{n+1}}<\frac{x_{n+1}}{y_{n+1}}<(a+\varepsilon)(1-\frac{y_{N+1}}{y_{n+1}})+\frac{x_{N+1}}{y_{n+1}}.$$

令$n\to\infty$分别取上，下极限，并注意到$y_n\to+\infty$,我们有

$$a-\varepsilon\le \lim\limits_{\overline{n\to\infty}}\frac{x_{n+1}}{y_{n+1}}\le \overline{\lim\limits_{n\to\infty}}\frac{x_{n+1}}{y_{n+1}}\le a+\varepsilon$$

由$\varepsilon>0$的任意性，我们有$a\le \lim\limits_\overline{n\to\infty}\frac{x_{n+1}}{y_{n+1}}\le \overline{\lim\limits_{n\to\infty}}\frac{x_{n+1}}{y_{n+1}}\le a$

因此$\lim\limits_{n\to\infty}\frac{x_n}{y_n}=\lim\limits_{\overline{n\to\infty}}\frac{x_n}{y_n}=\overline{\lim\limits_{n\to\infty}}\frac{x_n}{y_n}=a.$

stolz定理的证明