Question

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

• A. 1
• B. -1
• C. 0
• D. none of these

Answer 1

The correct option is A 1

$\underset{x\to \infty }{lim}\frac{\sqrt{x^2+1}-\sqrt[3]{3x^2+1}}{\sqrt[4]{4x^4+1}-\sqrt[5]{5x^4-1}}$
$=\underset{x\to \infty }{lim}\frac{\sqrt{1+1/x^2}-\sqrt[3]{31/x+1/x^3}}{\sqrt[4]{41+1/x^4}-\sqrt[5]{51/x-1/x^5}}=\frac{1-0}{1-0}=1$