Szczegółowe wyjaśnienie:
[tex]\lim\limits_{n\to\infty}\left(\dfrac{n^2+7}{n^2}\right)^{n^2-5}=\lim\limits_{n\to\infty}\dfrac{\left(\frac{n^2+7}{n^2}\right)^{n^2}}{\left(\frac{n^2+7}{n^2}\right)^5}=\dfrac{\lim\limits_{n\to\infty}\left(\frac{n^2+7}{n^2}\right)^{n^2}}{\lim\limits_{n\to\infty}\left(\frac{n^2+7}{n^2}\right)^5}=\dfrac{(1)}{(2)}[/tex]
[tex](1)\\\\\lim\limits_{n\to\infty}\left(\dfrac{n^2+7}{n^2}\right)^{n^2}=\lim\limits_{n\to\infty}\left(\dfrac{n^2}{n^2}+\dfrac{7}{n^2}\right)^{n^2}=\lim\limits_{n\to\infty}\left[\left(1+\dfrac{7}{n^2}\right)^{\frac{n^2}{7}}\right]^7=e^7[/tex]
[tex](2)\\\\\lim\limits_{n\to\infty}\left(\dfrac{n^2+7}{n^2}\right)^5=\lim\limits_{n\to\infty}\left(\dfrac{n^2\!\!\!\!\!\!\diagup\left(1+\frac{7}{n^2}\right)}{n^2\!\!\!\!\!\!\diagup}\right)^5=\lim\limits_{n\to\infty}\left(1+\dfrac{7}{n^2}\right)^5=(*)\\\\\dfrac{7}{n^2}\xrightarrow{n\to\infty}0\\\\(*)=\left(1+0\right)^5=1^5=1[/tex]
Z (1) i (2) mamy:
[tex]\lim\limits_{n\to\infty}\left(\dfrac{n^2+7}{n^2}\right)^{n^2-5}=\dfrac{e^7}{1}=e^7[/tex]
