Practicing Success
Let $f(x)=\left\{\begin{array}{cc}x^2\left|\cos \frac{\pi}{x}\right| & , x \neq 0 \\ 0 & , x=0\end{array}\right.$. Then, f is |
differentiable both at x = 0 and x = 2 differentiable at x = 0 but not differentiable at x = 2 not differentiable at x = 0 but differentiable at x = 2 differentiable neither at x = 0 nor at x = 2 |
differentiable at x = 0 but not differentiable at x = 2 |
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. |