Question

The value of $\underset{n\to \infty }{lim}\frac{\sqrt[4]{4n^5+2}-\sqrt[3]{3n^2+1}}{\sqrt[5]{5n^4+2}-\sqrt[2]{2n^3+1}}$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $\infty$

Answer 1

Answer: (B)

$0$

$\underset{n\to \infty }{lim}\frac{\sqrt[4]{4n^5+2}-\sqrt[3]{3n^2+1}}{\sqrt[5]{5n^4+2}-\sqrt[2]{2n^3+1}}$
$=\underset{n\to \infty }{lim}\frac{n^{\frac{8}{4}}\sqrt[4]{41+\frac{2}{n^5}}-n^{\frac{2}{3}}\sqrt[3]{31+\frac{1}{n^2}}}{n^{\frac{4}{5}}\sqrt[5]{51+\frac{2}{n^4}}-n^{\frac{3}{2}}\sqrt[2]{21+\frac{1}{n^3}}}$
$=\underset{n\to \infty }{lim}\frac{\frac{n^{\frac{5}{4}}}{n^{\frac{3}{2}}}\sqrt[4]{41+\frac{2}{n^5}}-\frac{n^{\frac{2}{3}}}{n^{\frac{5}{2}}}\sqrt[3]{31+\frac{1}{n^2}}}{\frac{n^{\frac{4}{5}}}{n^{\frac{3}{2}}}\sqrt[5]{51+\frac{2}{n^4}}-\frac{n^{\frac{3}{2}}}{n^{\frac{3}{2}}}\sqrt[2]{21+\frac{1}{n^3}}}$
$=\underset{n\to \infty }{lim}\frac{\frac{1}{n^{\frac{1}{4}}}\sqrt[4]{41+\frac{2}{n^5}}-\frac{1}{n^{\frac{5}{6}}}\sqrt[3]{31+\frac{1}{n^2}}}{\frac{1}{n^{\frac{7}{10}}}\sqrt[5]{51+\frac{2}{n^4}}-\sqrt[2]{21+\frac{1}{n^3}}}$
$=\frac{0-0}{0-1}=0$