If f(x) = \ cases x^ p ((1)/(x)), & x ≠ 0 0, & x = 0 cases ) is...

Given, f\left(x\right) = \left\{$\begin{array}{ll}{x}^{p}\cos \left(\dfrac{1}{x}\right), x \ne 0 \\ 0, x = 0\end{array}$\right)<br /><br />Since,f ( x )$is differentiable at$x = 0$, therefore it is continuous at$x = 0$<br /><br />$∴ \lim_{x \to 0}f\left(x\right) = f\left(0\right) = 0$<br /><br />$⇒ \lim_{x \to 0}{x}^{p}\cos \left(\dfrac{1}{x}\right) = 0 ⇒ p > 0$<br /><br />As$f\left(x\right)$is differentiable at$x = 0$<br /><br />$∴ \lim_{x \to 0}\dfrac{f\left(x\right) - f ( 0 )}{x - 0}$exists finitely<br /><br />$⇒ \lim_{x \to 0}\dfrac{{x}^{p}cos \dfrac{1}{x} - 0}{x}$exists finitely<br /><br />$⇒ \lim_{x \to 0}{x}^{p - 1}cos \dfrac{1}{x}$exists finitely<br /><br />$⇒ p - 1 > 0 ⇒ p > 1