Question

Let $f:R\to R$ be a function defined by
$f\left(x\right)=\{\begin{array}{cc}\frac{\sin \left(x^2\right)}{x}& \text{if}x\ne 0\\ 0& \text{if}x=0\end{array}$
Then, at $x=0$, $f$ is

• A. not continuous
• B. continuous but not differentiable
• C. differentiable and the derivative is not continuous
• D. differentiable and the derivative is continuous

Answer 1

The correct option is D differentiable and the derivative is continuous

For continuity,
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\sin \left(x^2\right)}{x}=\underset{x\to 0}{lim}\frac{\sin \left(x^2\right)}{x^2}\times x=0$
So the function is continuous at $x=0$
Checking the differentiability at $x=0$
Right hand derivative,
R.H.D $=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{\frac{\sin \left(h^2\right)}{h}-0}{h}=\underset{h\to 0}{lim}\frac{\sin \left(h^2\right)}{h^2}=1$
Left hand derivative,
L.H.D $=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{lim}\frac{\frac{\sin \left(\right(-h{)}^2)}{-h}-0}{-h}=\underset{h\to 0}{lim}\frac{\sin \left(h^2\right)}{h^2}=1$
So, L.H.D=R.H.D then function is differentiable at $x=0$ and $f^{\prime }\left(0\right)=1$
When $x\ne 0$
$f^{\prime }\left(x\right)=\frac{d\left(\frac{\sin \left(x^2\right)}{x}\right)}{dx}\Rightarrow f^{\prime }\left(x\right)=\frac{2x^2\cos \left(x^2\right)-\sin \left(x^2\right)}{x^2}\Rightarrow f^{\prime }\left(x\right)=2\cos \left(x^2\right)-\frac{\sin \left(x^2\right)}{x^2}{f}^{\prime }\left(x\right)=\{\begin{array}{cc}2\cos \left(x^2\right)-\frac{\sin \left(x^2\right)}{x^2}& \text{if}x\ne 0\\ 1& \text{if}x=0\end{array}$
Checking the continuity of $f^{\prime }\left(x\right)$,
$\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0}{lim}2\cos \left(x^2\right)-\frac{\sin \left(x^2\right)}{x^2}=2-1=1$
So the derivative is also continuous.