Question

If $f\left(x\right)=\frac{\sqrt{1+x}-\sqrt[3]{31+x}}{x}$ is continuous at $x=0$, then the value of $f\left(0\right)$ is

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

Answer 1

The correct option is A $\frac{1}{6}$

For continuity of $f\left(x\right)$ at $x=0$, we must have
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}\left(\frac{\sqrt{1+x}-\sqrt[3]{31+x}}{x}\right)$
$=\underset{x\to 0}{lim}\frac{(1+x{)}^{1/2}-(1+x{)}^{1/3}}{x}\left[\frac{0}{0}\text{form}\right]$
Using L Hospital's rule
$=\underset{x\to 0}{lim}\{\frac{1}{2(1+x{)}^{1/2}}-\frac{1}{3(1+x{)}^{2/3}}\}$
$=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$