Задача Кузнецов Пределы 4-2

Условие задачи

Вычислить предел числовой последовательности:

$\lim_{n\to \infty} n\left( {\sqrt {n\left( {n-2} \right)} -\sqrt {n^2-3}} \right)$

Решение

$\lim_{n\to \infty} n\left( {\sqrt {n\left( {n-2} \right)} -\sqrt {n^2-3}} \right) =$

$=\lim_{n\to \infty} \frac {n\left( {\sqrt {n\left( {n-2} \right)} -\sqrt {n^2-3}} \right)\left( {\sqrt {n\left( {n-2} \right)} +\sqrt {n^2-3}} \right)} {{\sqrt {n\left( {n-2} \right)} +\sqrt {n^2-3}}}=$

$=\lim_{n\to \infty} \frac {n\left(n\left(n-2\right) - \left(n^2-3\right)\right)} {{\sqrt {n\left( {n-2} \right)} +\sqrt {n^2-3}}} = \lim_{n\to \infty} \frac {n\left(n^2-2n - n^2+3\right)} {{\sqrt {n\left( {n-2} \right)} +\sqrt {n^2-3}}}=$

$= \lim_{n\to \infty} \frac {n\left(3 - 2n\right)} {{\sqrt {n\left( {n-2} \right)} +\sqrt {n^2-3}}}= \lim_{n\to \infty} \frac {3n - 2n^2} {{\sqrt {n\left( {n-2} \right)} +\sqrt {n^2-3}}}=$

$= \lim_{n\to \infty} \frac {\frac{1}{n}\left(3n - 2n^2\right)} {\frac{1}{n}\left({\sqrt {n\left( {n-2} \right)} +\sqrt {n^2-3}}\right)}=\lim_{n\to \infty} \frac {3 - 2n} {\sqrt {1 - \frac{2}{n}} + \sqrt {1-\frac{3}{n^2}}}=$

$=\left\{\frac{3-\infty}{\sqrt{1-0}+\sqrt{1-0}} = \frac{-\infty}{2}\right\}=-\infty$