Let  $f\left(x\right)=\left\{\begin{array}{cc}{x}^n\sin \frac{{\displaystyle 1}}{{\displaystyle x}},& x\ne 0\\ 0,& x=0\end{array}\right.$, then f(x) is continuous but not differentiable at x = 0 if $n\in (0,k] then k=$

since f(x) is continuous at x = 0, therefore 

$\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)=0\Rightarrow \underset{x\to 0}{\lim }{x}^n\sin \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)=0\Rightarrow n>0$

f(x) is differentiable at x = 0 if 

$\underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}$ exists finitely 

$\Rightarrow \underset{x\to 0}{\lim }\frac{x^n\sin \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)-0}{x}$ exists finitely 

$\Rightarrow \underset{x\to 0}{\lim }{x}^n\sin \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)existsfinitely\Rightarrow n-1>0\Rightarrow n>1$

 If  n$\le$1, then $\underset{x\to 0}{\lim }{x}^n\sin \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)$ does not exist and 

hence f(x) is not differentiable at x = 0 

hence f(x) is continuous but not differentiable at x = 0 for $0<n\le 1,i.en\in (0,1]$