Question

Find the following limit.
$\underset{n\to \infty }{lim}\frac{\sqrt[3]{34+-3n+n^6}}{1+3n-2n^2}.$

Answer 1

$\underset{n\to \infty }{lim}\frac{3\sqrt{4-3n+n^6}}{1+3n+2n^2}$

$=\underset{n\to \infty }{lim}\frac{(n^6{)}^{1/3}\left(\sqrt[3]{3\frac{4}{n^6}-\frac{3}{n^5}+1}\right)}{n^2(\frac{1}{n^2}+\frac{3}{n}-2)}$

$\Rightarrow \underset{n\to \infty }{lim}\frac{n^2\left(\sqrt[3]{3\frac{4}{n^6}-\frac{3}{n^5}+1}\right)}{n^2(\frac{1}{n^2}+\frac{3}{n}-2)}$

$=\underset{n\to \infty }{lim}\frac{\sqrt[3]{3\frac{4}{n^6}-\frac{3}{n^5}+1}}{\frac{1}{n^2}+\frac{3}{n}-2}$

$=\frac{-1}{2}$.