Question

Знайти границі послідовності.

Answer 1

Ответ:

$\displaystyle \frac{4}{25}$

Объяснение:

$\displaystyle \lim_{n\to \infty}(\frac{2n^2+n-1}{5n^2-7n+12})^2=\lim_{n\to \infty}(\frac{\frac{2n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}{\frac{5n^2}{n^2}-\frac{7n}{n^2}+\frac{12}{n^2}})^2=\\\\\lim_{n\to \infty}(\frac{2+\frac{1}{n}-\frac{1}{n^2}}{5-\frac{7}{n}+\frac{12}{n^2}})^2=(\frac{2}{5})^2=\frac{4}{25}$