Question

If function $f\left(x\right)=\frac{\sqrt{1+x}-\sqrt[3]{31+x}}{x},x\ne 0$ is continuous function, then $f\left(0\right)$ is equal to-

• A. $2$
• B. $\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$\frac{1}{6}$

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

We know that ${1+x}^n=1+nx+\frac{n(n-1)}{2}$+(Higher order terms)

Using this we can write $\sqrt{x+1}=1+\frac{x}{2}$and $\sqrt[3]{31+x}=1+\frac{x}{3}$

So $\sqrt{x+1}-\sqrt[3]{31+x}=\frac{x}{6}$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{x}{6x}=\frac{1}{6}$