Question

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

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

Answer 1

Answer: (A)

$1$

$\underset{x\to \infty }{lim}\frac{\sqrt{x^2+1}-\sqrt[3]{3x^2+1}}{\sqrt[4]{4x^4+1}-\sqrt[5]{5x^2+1}}$
$={lim}_{x\to \infty }\frac{x\sqrt{1+\frac{1}{x^2}}-x^{\frac{2}{3}}\sqrt[3]{31+\frac{1}{x^2}}}{x\sqrt{1+\frac{1}{x^4}}-x^{\frac{2}{5}}\sqrt[5]{51+\frac{1}{x^2}}}$
$=\underset{x\to \infty }{lim}\frac{x(\sqrt{1+\frac{1}{x^2}}-\frac{1}{x^{\frac{1}{3}}}\sqrt[3]{31+\frac{1}{x^2}})}{x(\sqrt{1+\frac{1}{x^4}}-\frac{1}{x^{\frac{3}{5}}}\sqrt[5]{51+\frac{1}{x^2}})}$
$=\frac{1}{1}=1$
Hence, option 'A' is correct.