Question

найти предел последовательности an=n^2-5n+1/3n-7

Answer 1

Ответ:

$n^2-5n+1/3n-7$

Пошаговое объяснение:

$\lim_{n \to \infty} a_n \displaystyle \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^2-5n+1}{3n-7}= \lim_{n \to \infty} \frac{\frac1{n^2}(n^2-5n+1)}{\frac1{n^2}(3n-7)}=\lim_{n \to \infty} \frac{1-\frac5n+\frac1{n^2}}{\frac3n-\frac7{n^2}}=\left[\frac{1-\frac5\infty+\frac1{\infty^2}}{\frac3\infty-\frac7{\infty^2}}\right] =\left[\frac{1-0+0}{0-0}\right]=\left[\frac10\right]=\infty$