Since, f(x) is continuous at x=0, therefore, limx→0f(x)=f(0)=0
⇒limx→0xnsin(1x)=0⇒n>0
f(x) is differential at x=0 if limx→0f(x)−f(0)x−0 exists finitely
⇒limx→0xnsin(1x)−0x exists finitely
⇒xn−1sin(1x) exists finitely.
⇒n−1>0⇒n>1
If n≤1, then limx→0xn−1sin(1x) does not exists and hence, f(x) is not differentiable at x=0
Hence, f(x) is continuous but not differentiable at x=0 for 0<n≤1
i.e., n∈(0,1]