Question

$\underset{x\to \infty }{lim}\frac{2\sqrt{x}+3\sqrt[3]{3x}+4\sqrt[4]{4x}+...+n\sqrt[n]{nx}}{\sqrt{(2x-3)}+\sqrt[3]{3(2x-3)}+\sqrt[4]{4(2x-3)}+...+\sqrt[n]{n(2x-3)}}$ is equal to

• A. $1$
• B. $\infty$
• C. $\sqrt{2}$
• D. None of these

Answer 1

The correct option is C $\sqrt{2}$

$\underset{x\to \infty }{lim}\frac{2\sqrt{x}+3\sqrt[3]{3x}+4\sqrt[4]{4x}+...+n\sqrt[n]{nx}}{\sqrt{(2x-3)}+\sqrt[3]{3(2x-3)}+\sqrt[4]{4(2x-3)}+...+\sqrt[n]{n(2x-3)}}$

Dividing numerator and denominator by $\sqrt{x}$

$=\underset{x\to \infty }{lim}\frac{2+3x^{-\frac{1}{6}}+4x^{-\frac{1}{4}}+...+n{x}^{\frac{(2-n)}{2n}}}{\sqrt{(2-\frac{3}{x})}+\frac{\sqrt[3]{3(2x-3)}}{\sqrt{x}}+...+\frac{\sqrt[n]{n(2x-3)}}{\sqrt{x}}}$

$=\underset{x\to \infty }{lim}\frac{2+\frac{3}{x^{1/6}}+\frac{4}{x^{1/4}}+...+\frac{n}{x^{\frac{(n-2)}{2n}}}}{\sqrt{(2-\frac{3}{x})}+\sqrt[6]{6\frac{{(2x-3)}^2}{x^3}}+...+\sqrt[2n]{2n\frac{{(2x-3)}^2}{x^n}}}$

$=\frac{2}{\sqrt{2}}=\sqrt{2}$