Question

Show that the function $f\left(x\right)=\left\{\begin{array}{ll}{x}^m\sin \left(\frac{1}{x}\right)& ,x\ne 0\\ 0& ,x=0\end{array}\right.$

(i) differentiable at x = 0, if m > 1
(ii) continuous but not differentiable at x = 0, if 0 < m < 1
(iii) neither continuous nor differentiable, if m ≤ 0

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}{x}^m\sin \left(\frac{1}{x}\right)\\ 0\end{array}\right.$ x≠0 , x=0

(i) Let m=2, then the function becomes $f\left(x\right)=\left\{\begin{array}{l}{x}^2\sin \left(\frac{1}{x}\right)\\ 0\end{array}\right.$ , x≠0, x=0

Differentiability at x=0:
$\underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)=0.$
[ ∵ $\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)=0$, as $\left|x\sin \frac{1}{x}-0\right|=\left|x\sin \frac{1}{x}\right|=\left|x\right|\left|\sin \frac{1}{x}\right|\le \left|x\right|$(∵$\left|\sin \theta \right|\le 1$ for all $\theta$) and hence $\left|x\sin \frac{1}{x}\right|<0$ when $\left|x-0\right|<\epsilon$$\left|x-0\right|<\epsilon$ ]
So, $f\text{'}\left(0\right)=0$, which means f is differentiable at x=0.
Hence the given function is differentiable at x=0.

(ii) Let $m=\frac{1}{2},0<m<1$. Then the function becomes
$f\left(x\right)=\left\{\begin{array}{l}{x}^{\frac{1}{2}}\\ 0\end{array}\sin \left(\frac{1}{x}\right)\right.$ , x≠0 , x=0

Continuity at x=0:
(LHL at x=0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(0-h)=\underset{h\to 0}{\lim }(-h{)}^{\frac{1}{2}}\sin \left(\frac{1}{0-h}\right)=\underset{h\to 0}{\lim }{h}^{\frac{1}{2}}\sin \left(\frac{1}{h}\right)=\underset{h\to 0}{\lim }{h}^{\frac{3}{2}}=0$.
(RHL at x=0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(0+h)=\underset{h\to 0}{\lim }{h}^{\frac{1}{2}}\sin \left(\frac{1}{h}\right)=\underset{h\to 0}{\lim }{h}^{\frac{3}{2}}=0$.
and $f\left(0\right)=0$
LHL at x=0 = RHL at x=0 = $\underset{x\to 0}{\lim }f\left(x\right)$,
Hence continuous.
Now Differentiabilty at x=0 when 0<m<1.
(LHD at x=0) = $\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{\lim }\frac{f(0-h)-f\left(0\right)}{0-h-0}=\underset{h\to 0}{\lim }\frac{(-h{)}^{{\displaystyle \frac{1}{2}}}\sin \left({\displaystyle \frac{1}{-h}}\right)}{-h}$