Question

The value of $\underset{x\to \infty }{lim}\frac{2\sqrt{x}+3\sqrt[3]{3x}+5\sqrt[5]{5x}}{\sqrt{3x+2}+\sqrt[3]{32x-3}}=$

• A. $\frac{2}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. none of these

Answer 1

Answer: (A)

$\frac{2}{\sqrt{3}}$

Let, $L=\underset{x\to \infty }{lim}\frac{2\sqrt{x}+3\sqrt[3]{3x}+5\sqrt[5]{5x}}{\sqrt{3x+2}+\sqrt[3]{32x-3}}$
Divide numerator and denominator by $\sqrt{x}$
$L=\underset{x\to \infty }{lim}\frac{2+3\sqrt[3]{31/x}+5\sqrt[5]{51/x^3}}{\sqrt{3+2/x}+\sqrt[3]{32/x-3/x^3}}=\frac{2}{\sqrt{3}}$