## Problem Statement

Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \sqrt{n^{3}+8}\left(\sqrt{n^{3}+2}-\sqrt{n^{3}-1}\right)$
## Solution

$\lim _{n \rightarrow \infty} \sqrt{n^{3}+8}\left(\sqrt{n^{3}+2}-\sqrt{n^{3}-1}\right)=$

$=\lim _{n \rightarrow \infty} \frac{\sqrt{n^{3}+8}\left(\sqrt{n^{3}+2}-\sqrt{n^{3}-1}\right)\left(\sqrt{n^{3}+2}+\sqrt{n^{3}-1}\right)}{\sqrt{n^{3}+2}+\sqrt{n^{3}-1}}=$

$=\lim _{n \rightarrow \infty} \frac{\sqrt{n^{3}+8}\left(n^{3}+2-\left(n^{3}-1\right)\right)}{\sqrt{n^{3}+2}+\sqrt{n^{3}-1}}=\lim _{n \rightarrow \infty} \frac{3 \sqrt{n^{3}+8}}{\sqrt{n^{3}+2}+\sqrt{n^{3}-1}}=$

$=\lim _{n \rightarrow \infty} \frac{\frac{1}{n} 3 \sqrt{n^{3}+8}}{\frac{1}{n}\left(\sqrt{n^{3}+2}+\sqrt{n^{3}-1}\right)}=\lim _{n \rightarrow \infty} \frac{3 \sqrt{1+\frac{8}{n^{3}}}}{\sqrt{1+\frac{2}{n^{3}}}+\sqrt{1-\frac{1}{n^{3}}}}=$

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

## Problem Kuznetsov Limits 5-19