Question

Let, $\{\begin{array}{cc}x\times \sin \left(\frac{1}{x}\right)& ifx\ne 0\\ 1& ifx=0,then\end{array}$

• A. f(x) is not continuous at x = 0
• B. f(x) is continous at x = 0, but not differentiabble at x = 0
• C. f(x)is differentiable at x = 0 and f(0) = 0
• D. f(x) is differentiable at x = 0 and f(0) = 1

Answer 1

The correct option is B f(x) is continous at x = 0, but not differentiabble at x = 0

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

Which does not exists as it oscillates between -1 and 1.

Similarly, $L{f}^{\prime }\left(0\right)$ does not exists. Hence, f(x) is not differtiable at x=0

For continuity,
$f(0+h)=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}h\times \sin \left(\frac{1}{h}\right)=0$
And, $f(0-h)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}\{-h\times \sin \left(\frac{1}{-h}\right)\}=0$

Also, $f\left(0\right)=0\left(given\right)$

So, the function, f(x) is continous at x = 0 but is not differentiable at x = 0.