Question

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

• A. 1
• B. 0
• C. -1
• D. $\infty$

Answer 1

The correct option is B

0

$\underset{n\to \infty }{lim}\frac{\sqrt[4]{4n^5+2}-\sqrt[3]{3n^2+1}}{\sqrt[5]{5n^4+2}-\sqrt[2]{2n^3+1}}=\underset{n\to \infty }{lim}\frac{n^{5/4}\sqrt[4]{41+\frac{2}{n^5}}-n^{2/3\sqrt[3]{31+\frac{1}{n^2}}}}{n^{4/5}\sqrt[5]{51+\frac{2}{n^4}-n^{3/2\sqrt[2]{21+\frac{1}{n^3}}}}}=\underset{n\to \infty }{lim}\frac{\frac{n^{5/4}}{n^{3/2}}\sqrt[4]{41+\frac{2}{n^5}}-\frac{n^{2/3}}{n^{3/2}}\sqrt[3]{31+\frac{1}{n^2}}}{\frac{n^{4/5}}{n^{3/2}}\sqrt[5]{51+\frac{2}{n^4}}-\frac{n^3/2}{n^{3/2}}\sqrt[2]{21+\frac{1}{n^3}}}{\text{On dividing the numerator and denominator by the highest power of n.i.e., n}}^{3/2}=\underset{n\to \infty }{lim}\frac{\frac{1}{n^{1/4}}\sqrt[4]{41+\frac{2}{n^5}}-\frac{1}{n^{5/6}}\sqrt[3]{31+\frac{1}{n^2}}}{n^{7/10}\sqrt[5]{51+\frac{2}{n^4}}-\sqrt[2]{21+\frac{1}{n^3}}}=\frac{0-0}{0-1}=0$