If $f(x)= \begin{cases}x^m \sin \left(\frac{1}{x}\right), & x \neq 0 \\ ~~~~~~0, & x=0\end{cases}$ is continuous at x = 0, then

$m \in(0, \infty)$

If f(x) is continuous at x = 0, then

$\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)=0$

Now,

$\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{h \rightarrow 0} f(0-h)=\lim\limits_{h \rightarrow 0}(-h)^m \sin \left(-\frac{1}{h}\right)$

$\Rightarrow \lim\limits_{x \rightarrow 0^{-}} f(x)=-\lim\limits_{h \rightarrow 0}(-h)^m \sin \left(\frac{1}{h}\right)=0$, only when m > 0

and, $\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{h \rightarrow 0} f(0+h)$

$\Rightarrow \lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0} h^m \sin \left(\frac{1}{h}\right)=0$, only when m > 0

Hence, f(x) is continuous at x = 0, if m > 0