Question

The value of $f\left(0\right)$ so that $f\left(x\right)=\frac{\sqrt{1+x}-\sqrt[3]{31+x}}{x}$ is continuous is

• A. $\frac{1}{6}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $-1$

Answer 1

Answer: (A)

$\frac{1}{6}$

Given: $f\left(x\right)=\frac{\sqrt{1+x}-\sqrt[3]{31+4}}{x}$

To find: Value of $f\left(0\right)$

Sol: ${lim}_{x\to 0}\frac{{(1+x)}^{\frac{1}{2}}-{(1+x)}^{\frac{1}{3}}}{x}$

By binomial expansion:

${(1+x)}^{\frac{1}{2}}=1+\frac{1}{2}x{+}^{\frac{1}{2}}{C}_2.x^2+.....$ and

${(1+x)}^{\frac{1}{3}}=1+\frac{1}{3}x{+}^{\frac{1}{3}}{C}_2.x^2+.....$

$⟹\frac{{(1+x)}^{\frac{1}{2}}-{(1+x)}^{\frac{1}{3}}}{x}=\frac{1}{6}+{(}^{\frac{1}{2}}{C}_2{-}^{\frac{1}{3}}{C}_2)x^2+.....$ higher order of x

$⟹{lim}_{x\to 0}\frac{{(1+x)}^{\frac{1}{2}}-{(1+x)}^{\frac{1}{3}}}{x}=\frac{1}{6}+0$

Hence ,correct answer is $\frac{1}{6}$