Задача Кузнецов Пределы 4-31

Условие задачи

Вычислить предел числовой последовательности:

$\lim_{n\to \infty} \frac{\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} - \sqrt{n^6-3n^3+5}} {n}$

Решение

$\lim_{n\to \infty} \frac{\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} - \sqrt{n^6-3n^3+5}} {n}=$

$=\lim_{n\to \infty} \frac{\left(\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} - \sqrt{n^6-3n^3+5}\right)\left(\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} + \sqrt{n^6-3n^3+5}\right)} {n\left(\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} + \sqrt{n^6-3n^3+5}\right)}=$

$=\lim_{n\to \infty} \frac{\left(n^2+5 \right)\left(n^4+2 \right) - \left(n^6-3n^3+5\right)} {n\left(\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} + \sqrt{n^6-3n^3+5}\right)}=$

$=\lim_{n\to \infty} \frac{n^6+5n^4+2n^2+10 -n^6+3n^3-5} {n\left(\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} + \sqrt{n^6-3n^3+5}\right)}=$

$=\lim_{n\to \infty} \frac{5n^4+3n^3+2n^2+5} {n\left(\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} + \sqrt{n^6-3n^3+5}\right)}=$

$=\lim_{n\to \infty} \frac{\frac{1}{n^4}\left(5n^4+3n^3+2n^2+5\right)} {\frac{1}{n^4}n\left(\sqrt{\left(n^2+5 \right)\left(n^4+2 \right)} + \sqrt{n^6-3n^3+5}\right)}=$

$=\lim_{n\to \infty} \frac{5+\frac{3}{n}+\frac{2}{n^2}+\frac{5}{n^4}} {\sqrt{\left(1+\frac{5}{n^2} \right)\left(1+\frac{2}{n^4} \right)} + \sqrt{1-\frac{3}{n^3}+\frac{5}{n^6}}}=$

$= \frac{5+0+0+0} {\sqrt{(1+0)(1+0)} + \sqrt{1-0+0}}=\frac{5}{2}$