Odpowiedź:
Granicą tego ciągu jest 4/5
Szczegółowe wyjaśnienie:
Z twierdzenia o trzech ciągach:
[tex]\lim\limits_{n\to\infty}\sqrt[n]{0+0+\left(\dfrac{4}{5}\right)^n} \leq \lim\limits_{n\to\infty}\sqrt[n]{\left(\dfrac{2}{5}\right)^n+\left(\dfrac{1}{4}\right)^n+\left(\dfrac{4}{5}\right)^n}\leq \lim\limits_{n\to\infty}\sqrt[n]{\left(\dfrac{4}{5}\right)^n+\left(\dfrac{4}{5}\right)^n+\left(\dfrac{4}{5}\right)^n}[/tex]
[tex]\lim\limits_{n\to\infty}\dfrac{4}{5}\sqrt[n]{1} \leq \lim\limits_{n\to\infty}\sqrt[n]{\left(\dfrac{2}{5}\right)^n+\left(\dfrac{1}{4}\right)^n+\left(\dfrac{4}{5}\right)^n}\leq \lim\limits_{n\to\infty}\dfrac{4}{5}\sqrt[n]{3}[/tex]
[tex]\dfrac{4}{5} \leq \lim\limits_{n\to\infty}\sqrt[n]{\left(\dfrac{2}{5}\right)^n+\left(\dfrac{1}{4}\right)^n+\left(\dfrac{4}{5}\right)^n}\leq \dfrac{4}{5}[/tex]
[tex]\lim\limits_{n\to\infty}\sqrt[n]{\left(\dfrac{2}{5}\right)^n+\left(\dfrac{1}{4}\right)^n+\left(\dfrac{4}{5}\right)^n}= \dfrac{4}{5}[/tex]
