Question

The value of $f\left(0\right)$ so that the function
$f\left(x\right)=\frac{\sqrt{1+x}-{(1+x)}^{1/3}}{x}$ becomes continuous is equal to-

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

Answer 1

Answer: (A)

$\frac{1}{6}$

Given $f\left(x\right)=\frac{\sqrt{(1+x)}-{(1+x)}^{1/3}}{x}$
Since,f(x) is continuous at $x=0$
$f\left(0\right)=LHL=RHL$
$RHL=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)$
$=\underset{h\to 0}{lim}\frac{{(1+h)}^{1/2}-{(1+h)}^{1/3}}{h}$
$=\frac{(1+\frac{1}{2}h-\frac{1}{8}{h}^2.....)-(1+\frac{h}{3}-\frac{1}{9}{h}^2+....)}{h}$
$=\frac{(\frac{h}{2}-\frac{h}{3})+higherdegreetermsofh}{h}$
$=\underset{h\to 0}{lim}\frac{h[(\frac{1}{2}-\frac{1}{3})+higherdegreetermsofh]}{h}$
$RHL=\frac{1}{6}+0=\frac{1}{6}$
Hence, $f\left(0\right)=\frac{1}{6}$