Odpowiedź:
[tex]\huge\boxed{a_n\xrightarrow{n\to\infty}\dfrac{1}{4};\ b_n\xrightarrow{n\to\infty}\dfrac{7}{64}}[/tex]
Szczegółowe wyjaśnienie:
[tex]\lim\limits_{n\to\infty}\dfrac{n^2-3n+2}{4n^2+2n-1}=\lim\limits_{n\to\infty}\dfrac{n^2\!\!\!\!\!\!\diagup\left(1-\frac{3}{n}+\frac{2}{n^2}\right)}{n^2\!\!\!\!\!\!\diagup\left(4+\frac{2}{n}-\frac{1}{n^2}\right)}=\dfrac{1}{4}[/tex]
Wyjaśnienie:
[tex]n\to\infty\ \text{wtedy}\ -\dfrac{3}{n}\to0;\ \dfrac{2}{n^2}\to0;\ \dfrac{2}{n}\to0;\ -\dfrac{1}{n}\to0[/tex]
[tex]\lim\limits_{n\to\infty}\dfrac{7n^3(n-2)^2}{(n-2)(8n^2+2)^2}=\lim\limits_{n\to\infty}\dfrac{7n^3(n-2)}{(8n^2+2)^2}=\lim\limits_{n\to\infty}\dfrac{7n^4-14n^3}{64n^4+32n^2+4}\\\\=\lim\limits_{n\to\infty}\dfrac{n^4\!\!\!\!\!\!\diagup\left(7-\frac{14}{n}\right)}{n^4\!\!\!\!\!\!\diagup\left(64+\frac{32}{n^2}+\frac{4}{n^2}\right)}=\dfrac{7}{64}[/tex]
Wyjaśnienie:
[tex]n\to\infty\ \text{wtedy}\ -\dfrac{14}{n}\to0;\ \dfrac{32}{n^2}\to0;\ \dfrac{4}{n^4}\to0[/tex]