If $f(x)=\left\{\begin{array}{cc}(x-a)^n \cos \left(\frac{1}{x-a}\right), & x \neq a \\ 0, & x=a\end{array}\right.$ then at x = a, f(x) is

continuous if n > 0 and differentiable if n > 1

For f(x) to be continuous at x = a, we must have

$\lim\limits_{x \rightarrow a} f(x)=f(a)$

$\Rightarrow \lim\limits_{x \rightarrow a}(x-a)^n \cos \left(\frac{1}{x-a}\right)=0$

$\Rightarrow n>0$

For f(x) to be differentiable

$\lim\limits_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$  must exist finitely.

$\Rightarrow \lim\limits_{x \rightarrow a} \frac{(x-a)^n \cos \left(\frac{1}{x-a}\right)-0}{x-a}$ must exist finitely

$\Rightarrow \lim\limits_{x \rightarrow a}(x-a)^{n-1} \cos \left(\frac{1}{x-a}\right)$ must exist finitely

$\Rightarrow n-1>0$

$\Rightarrow n>1$

Hence, f(x) is continuous for n > 0 and differentiable for n > 1.