If $f(x)=\left\{\begin{matrix}x^a.\sin(\frac{1}{x}),&x≠1\\0,&x=0\end{matrix}\right.$ a is continuous but non-differentiable at x = 0, then

a ∈ (0, 1]

$f'(0)=\underset{h→0}{\lim}\frac{f(h)-f(0)}{h}=\underset{h→0}{\lim}\frac{h^a\sin(\frac{1}{h})}{h}=\underset{h→0}{\lim}h^{a-1}.\sin(\frac{1}{h})$

This limit will not exist if $a-1≤0⇒a≤1$