Question

If $f x=\sqrt{1-\sqrt{1-x^2}}$, then at $x=0$

• A. $f\left(x\right)$ is differentiable as well as continuous
• B. $f\left(x\right)$ is differentiable but not continuous
• C. $f\left(x\right)$ is continuous but not differentiable
• D. $f\left(x\right)$ is neither continuous nor differentiable

Answer 1

Answer: (C)

$L f^{\prime}(0)=\displaystyle\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\sqrt{1-\sqrt{1-h^2}}}{-h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\sqrt{1-\sqrt{1-h^2}}}{-h} \times \frac{\sqrt{1+\sqrt{1-h^2}}}{\sqrt{1+\sqrt{1-h^2}}} $
$=\displaystyle\lim _{h \rightarrow 0} \frac{-1}{\sqrt{1+\sqrt{1-h^2}}}=\frac{-1}{\sqrt{2}}$
$R f^{\prime}(0)=\displaystyle\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\displaystyle\lim _{h \rightarrow 0} \frac{\sqrt{1-\sqrt{1-h^2}}}{h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{1}{\sqrt{1+\sqrt{1-h^2}}}=\frac{1}{\sqrt{2}} $
Therefore, $f(x)$ is not differentiable at $x=0$.
since $L f^{\prime}(0)$ and $R f^{\prime}(0)$ are finite therefore, us at $x=0$.