Odpowiedź + Szczegółowe wyjaśnienie:
[tex]a)\ \lim\limits_{n\to\infty}(10n^5+n^2-1)=\lim\limits_{n\to\infty}n^5\left(10+\dfrac{1}{n^3}-\dfrac{1}{n^5}\right)=\infty[/tex]
[tex]\dfrac{1}{n^3}\xrightarrow{x\to\infty}0\\\\\dfrac{1}{n^5}\xrightarrow{n\to\infty}0[/tex]
[tex]b)\ \lim\limits_{n\to\infty}\dfrac{n^6-n^8}{n^5+2n^8}=\lim\limits_{n\to\infty}\dfrac{n^8\!\!\!\!\!\!\diagup\left(\frac{1}{n^2}-1\right)}{n^8\!\!\!\!\!\!\diagup\left(\frac{1}{n^3}+2\right)}=-\dfrac{1}{2}\\\\\dfrac{1}{n^2}\xrightarrow{n\to\infty}0\\\\\dfrac{1}{n^3}\xrightarrow{n\to\infty}0[/tex]
[tex]c)\ \lim\limits_{n\to\infty}\dfrac{4n^7+2n^5}{n^6+6n^8}=\lim\limits_{n\to\infty}\dfrac{n^8\!\!\!\!\!\!\diagup\left(\frac{4}{n}+\frac{2}{n^3}\right)}{n^8\!\!\!\!\!\!\diagup\left(\frac{1}{n^2}+6\right)}=\dfrac{0}{6}=0\\\\\dfrac{4}{n}\xrightarrow{n\to\infty}0\\\\\dfrac{2}{n^3}\xrightarrow{n\to\infty}0\\\\\dfrac{1}{n^2}\xrightarrow{n\to\infty}0[/tex]
[tex]d)\ \lim\limits_{n\to\infty}\dfrac{n^5-n^9}{n^4+n^7}=\lim\limits_{n\to\infty}\dfrac{n^7\!\!\!\!\!\!\diagup\left(\frac{1}{n^2}-n^2\right)}{n^7\!\!\!\!\!\!\diagup\left(\frac{1}{n^3}+1\right)}=\dfrac{-\infty}{1}=-\infty\\\\\dfrac{1}{n^2}\xrightarrow{n\to\infty}0\\\\\dfrac{1}{n^3}\xrightarrow{n\to\infty}0\\\\n^2\xrightarrow{n\to\infty}\infty[/tex]
