Let f(x)=≤ft […] .. Then, f is

We observe that

$\lim\limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim\limits_{x \rightarrow 0} \frac{x^2\left|\cos \frac{\pi}{x}\right|-0}{x-0}=\lim\limits_{x \rightarrow 0} x\left|\cos \frac{\pi}{x}\right|=0$

So, f(x) is differentiable at x = 0.

Now,

$\lim\limits_{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}$

$=\lim\limits_{h \rightarrow 0} \frac{f(2-h)-f(2)}{(2-h)-2}$

$=\lim\limits_{h \rightarrow 0} \frac{(2-h)^2\left|\cos \left(\frac{\pi}{2-h}\right)\right|}{- ज}$

$=\lim\limits_{h \rightarrow 0} \frac{-(2-h)^2 \cos \left(\frac{\pi}{2-h}\right)}{-h}$

$=\lim\limits_{h \rightarrow 0} \frac{(2-h)^2 \sin \left(\frac{\pi}{2}-\frac{\pi}{2-h}\right )}{h}$

$=\lim\limits_{h \rightarrow 0} \frac{(2-h)^2}{h} \sin \left\{\frac{-\pi h}{2(2-h)}\right \}$

$=-\lim\limits_{h \rightarrow 0} \frac{\sin \left\{\frac{\pi h}{2(2-h)}\right\}}{\frac{\pi h} {2(2-h)}} \times \frac{(2-h) \pi}{2}=-\pi$

and,

$\lim\limits_{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}$

$= \lim\limits_{h \rightarrow 0} \frac{f(2+h)-f(2)}{(2+h)-2} $

$= \lim\limits_{h \rightarrow 0} \frac{(2+h)^2 \cos \left(\frac{\pi}{2+h}\right)-0}{h}$

$= \lim\limits_{h \rightarrow 0} \frac{(2+h)^2}{h} \sin \left(\frac{\pi}{2}-\frac{\pi}{2+h}\right)$

$=\lim\limits_{h \rightarrow 0} \frac{(2+h)^2}{h} \sin \left\{\frac{\pi h}{2(2+h)}\right\}$

$=\lim\limits_{h \rightarrow 0} \frac{\sin \left\{\frac{\pi h}{2(2+h)}\right\}}{\frac{\pi h}{2(2+h)}} \times \frac{\pi}{2}(2+h)=\frac{\pi}{2} \times 2=\pi$

∴  $\lim\limits_{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2} \neq \lim\limits_{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}$

So, f(x) is not differentiable at x = 2.