Odpowiedź:
[tex]\lim\limits_{x\to\infty}\dfrac{2n^3-4n-1}{6n+3n^2-n^3}=-2[/tex]
Szczegółowe wyjaśnienie:
[tex]\lim\limits_{x\to\infty}\dfrac{2n^3-4n-1}{6n+3n^2-n^3}=\lim\limits_{x\to\infty}\dfrac{n^3(2-\frac{4}{n^2}-\frac{1}{n^3})}{n^3(\frac{6}{n^2}+\frac{3}{n}-1)}=\\\\\lim\limits_{x\to\infty}\dfrac{2-\frac{4}{n^2}-\frac{1}{n^3}}{\frac{6}{n^2}+\frac{3}{n}-1}=\dfrac{2-0-0}{0+0-1}=\dfrac{2}{-1}=\boxed{-2}[/tex]
