Mamy obliczyć granicę ciągu:
[tex]a_n=\left(\dfrac{n^2+2}{2n^2+1}\right)^{n^2}[/tex]
[tex]\lim\limits_{n\to\infty}\left(\dfrac{n^2+2}{2n^2+1}\right)^{n^2}=\lim\limits_{n\to\infty}\left(\dfrac{n^2\left(1+\frac{2}{n^2}\right)}{n^2\left(2+\frac{1}{n^2}\right)}\right)^{n^2}=\lim\limits_{n\to\infty}\left(\dfrac{1+\frac{2}{n^2}}{2+\frac{1}{n^2}}\right)^{n^2}[/tex]
Jako, że:
[tex]\dfrac{1}{n^2}\xrightarrow{n\to\infty}0\\\\\dfrac{2}{n^2}\xrightarrow{n\to\infty}0[/tex]
otrzymujemy
[tex]=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\right)^{n^2}=0[/tex]
Ostatecznie:
[tex]\huge\boxed{\lim\limits_{n\to\infty}\left(\dfrac{n^2+2}{2n^2+1}\right)^{n^2}=0}[/tex]
