Question

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

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

Answer 1

Answer: (B)

$0$

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

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

because $\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3},\frac{1}{x^4}$ and $\frac{1}{x^5}$ all approach 0 as $x\to \infty$.