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. $-2$

Answer 1

Answer: (A)

$1$

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

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

$=\frac{1}{1}$

$=1$