Question

If $f\left(x\right)=\frac{1}{x}\sin x^2,x\ne 0,f\left(0\right)=0,$ discuss the continuity and differentiability of $f\left(x\right)$ at $x=0$.

• A. $f\left(x\right)$ is continuous and differentiable and f'(0)$=1$
• B. $f\left(x\right)$ is discontinuous and differentiable
• C. $f\left(x\right)$ is continuous and not-differentiable
• D. $f\left(x\right)$ is neither continuous nor differentiable

Answer 1

The correct option is A $f\left(x\right)$ is continuous and differentiable and f'(0)$=1$

$f\left(0\right)=0$

$f(0+0)=\underset{h\to 0}{lim}\frac{\sin {(0+h)}^2}{(0+h)}$ $=\underset{h\to 0}{lim}\frac{\sin h^2}{h^2}.h=1\times 0=0.$

And $f(0-0)=\underset{h\to 0}{lim}\frac{\sin {(0-h)}^2}{0-h}.$ $=\underset{h\to 0}{lim}\frac{\sin h^2}{h^2}(-h)=1\times 0=0.$

Hence $f\left(x\right)$ is continuous at $x=0$

Now $R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}(\frac{\sin {(0+h)}^2}{0+h}-0)/h=\underset{h\to 0}{lim}\frac{\sin h^2}{h^2}=1$

$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\left(\frac{\sin (0+h-0)}{0+h}\right)/(-h)=\underset{h\to 0}{lim}\frac{\sin h^2}{h^2}=1$

Hence $f\left(x\right)$ is differentiable at $x=0and{f}^{\prime }\left(0\right)=1.$
Ans: A