Question

If $fx=\sqrt{1-\sqrt{1-x^2}}$, 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)

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