Question

$\underset{x\to 0}{lim}\frac{\sqrt[3]{31+x^2}-\sqrt[4]{41+2x}}{x+x^2}$ is equal to

• A. $\frac{1}{4}$
• B. $-\frac{1}{2}$
• C. $1$
• D. Does not exist

Answer 1

The correct option is B $-\frac{1}{2}$

We have,

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

This is $\frac{0}{0}$ form, so apply L-Hospital rule

$\underset{x\to 0}{lim}\frac{\frac{1}{3}{(1+x^2)}^{-\frac{2}{3}}\times (0+2x)-\frac{1}{4}{(1+2x)}^{-\frac{3}{4}}\times (0+2)}{1+2x}$

$\underset{x\to 0}{lim}\frac{\frac{2x}{3{(1+x^2)}^{\frac{2}{3}}}-\frac{1}{2{(1+2x)}^{\frac{3}{4}}}}{1+2x}$

$\Rightarrow \frac{\frac{0}{3{(1+0)}^{\frac{2}{3}}}-\frac{1}{2{(1+0)}^{\frac{3}{4}}}}{1+0}$

$\Rightarrow -\frac{1}{2{\left(1\right)}^{\frac{3}{4}}}$

$\Rightarrow -\frac{1}{2}$

Hence, this is the answer.