Since at $x = 0 , f(x)$ is a continuous function
$\therefore $ $\mathop {\lim }\limits_{x \to 0} f(x) = f(0) = 0$
==> $\mathop {\lim }\limits_{x \to 0} \,{x^p}\sin \frac{1}{x} = 0 \Rightarrow p > 0$.
$f(x)$ is differentiable at $x = 0$, if $\mathop {\lim }\limits_{x \to 0} \frac{{f(x) - f(0)}}{{x - 0}}$ exists
==> $\mathop {\lim }\limits_{x \to 0} \frac{{{x^p}\sin \frac{1}{x} - 0}}{{x - 0}}$ exists
==> $\mathop {\lim }\limits_{x \to 0} {x^{p - 1}}\sin \frac{1}{x}$ exists
==> $p - 1 > 0$ or $p > 1$
If $p \le 1$, then $\mathop {\lim }\limits_{x \to 0} {x^{p - 1}}\sin \left( {\frac{1}{x}} \right)$ does not exist and at $x = 0$ $f(x)$ is not differentiable.
$\therefore $ for $0 < p \le 1$ $f(x)$ is a continuous function at $x = 0$ but not differentiable.
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