Let f(x) = \ array ll x^ p sin(1)/(x), x ≠ 0 array cases 0, & x = 0...

We have,

$f$\left(x\right) = \left\{$$\begin{array}{ll}{x}^{p}sin\dfrac{1}{x}, x \ne 0\end{array}$$ $\begin{cases}0, & x = 0\end{cases}$\right)

For $f\left(x\right)$ to be continuous at $x = 0$

$\lim_{x \to 0}f\left(x\right) = f\left(0\right)$

$⇒ \lim_{x \to 0}\left({x}^{p}sin\dfrac{1}{x}\right) = 0$

$⇒ \lim_{x \to 0}\left({x}^{p}\right) \times \lim_{x \to 0}\left(sin\dfrac{1}{x}\right) = 0$

$⇒ \lim_{x \to 0}\left({x}^{p}\right) \times \left(\text{value between} - 1\text{to}1\right) = 0$

$∴ p > 0$

Now,

RHD at $\text{RHD at}\left(x = 0\right) = \lim_{h \to 0}\left[\dfrac{{h}^{p}sin\dfrac{1}{h} - 0}{h}\right] = \lim_{h \to 0}\left[{h}^{p - 1}sin\dfrac{1}{h}\right]$

$\text{LHD at}\left(x = 0\right) = \lim_{h \to 0}\left[\dfrac{{\left(- h\right)}^{p}\sin \left(- \dfrac{1}{h}\right) - 0}{- h}\right] = \lim_{h \to 0}\left[{\left(- 1\right)}^{p}{\left(- h\right)}^{p - 1}\sin \left(\dfrac{1}{h}\right)\right]$

Since, $f\left(x\right)$ is not differentiable at $x = 0$, therefore

$p \le 1$

Hence, $0 < p \le 1$.