Odpowiedź:
[tex]\huge\boxed{\lim\limits_{n\to\infty}\dfrac{(3n+1)^2(n-1)^2}{(3-2n^2)^2}=\dfrac{9}{4}}[/tex]
Szczegółowe wyjaśnienie:
[tex]\lim\limits_{n\to\infty}\dfrac{(3n+1)^2(n-1)^2}{(3-2n^2)^2}=\lim\limits_{n\to\infty}\dfrac{(9n^2+6n+1)(n^2-2n+1)}{9-12n^2+4n^4}\\\\=\lim\limits_{n\to\infty}\dfrac{9n^4-18n^3+9n^2+6n^3-12n^2+6n+n^2-2n+1}{9-12n^2+4n^4}\\\\=\lim\limits_{n\to\infty}\dfrac{9n^4-12n^3-2n^2+4n+1}{9-12n^2+4n^4}=\lim\limits_{n\to\infty}\dfrac{n^4\!\!\!\!\!\!\diagup\left(9-\frac{12}{n}-\frac{2}{n^2}+\frac{4}{n^3}+\frac{1}{n^4}\right)}{n^4\!\!\!\!\!\!\diagup\left(\frac{9}{n^4}-\frac{12}{n^2}+4\right)}=\dfrac{9}{4}[/tex]
[tex]-\dfrac{12}{n}\xrightarrow{n\to\infty}0\\\\-\dfrac{2}{n^2}\xrightarrow{n\to\infty}0\\\\\dfrac{4}{n^3}\xrightarrow{n\to\infty}0\\\\\dfrac{1}{n^4}\xrightarrow{n\to\infty}0\\\\\dfrac{9}{n^4}\xrightarrow{n\to\infty}0\\\\-\dfrac{12}{n^2}\xrightarrow{n\to\infty}0[/tex]