Szczegółowe wyjaśnienie:
[tex]\lim\limits_{n\to\infty}\left(\dfrac{n}{n-9}\right)^{n+1}=\lim\limits_{n\to\infty}\left(\dfrac{n-9+9}{n-9}\right)^{n+1}=\lim\limits_{n\to\infty}\left(\dfrac{n-9}{n-9}+\dfrac{9}{n-9}\right)^{n+1}\\\\=\lim\limits_{n\to\infty}\left(1+\dfrac{9}{n-9}\right)^{n+1}=\lim\limits_{n\to\infty}\left\{\left[\left(1+\dfrac{9}{n-9}\right)^{\frac{n-9}{9}}\right]^{\frac{9}{n-9}}\right\}^{n+1}[/tex]
[tex]=\lim\limits_{n\to\infty}\left[\left(1+\dfrac{9}{n-9}\right)^{\frac{n-9}{9}}\right]^{\frac{9}{n-9}\cdot(n+1)}=\lim\limits_{n\to\infty}\left[\left(1+\dfrac{9}{n-9}\right)^{\frac{n-9}{9}}\right]^{\frac{9n+9}{n-9}}\\\\=\lim\limits_{n\to\infty}\left[\left(1+\dfrac{9}{n-9}\right)^{\frac{n-9}{9}}\right]^{\frac{n\left(9+\frac{9}{n}\right)}{n\left(1-\frac{9}{n}\right)}}=\lim\limits_{n\to\infty}\left[\left(1+\dfrac{9}{n-9}\right)^{\frac{n-9}{9}}\right]^{\frac{9+\frac{9}{n}}{1-\frac{9}{n}}}\\\\=e^{\frac{9}{1}}=e^9[/tex]
[tex]\left(1+\dfrac{9}{n-9}\right)^{\frac{n-9}{9}}\xrightarrow{n\to\infty} e\\\\\dfrac{9}{n}\xrightarrow{n\to\infty}0\\\\\dfrac{1}{n}\xrightarrow{n\to\infty}0\\\\\dfrac{9+\frac{9}{n}}{1-\frac{9}{n}}\xrightarrow{n\to\infty}\dfrac{9+0}{1-0}=\dfrac{9}{1}=9[/tex]
Przy takich granicach dążymy do postaci:
[tex]\left(1+\dfrac{1}{n}\right)^{n}[/tex]
