f(x) = $x^2$ sin$\frac{1}{x}$, f(0) = 0

Continuity at 0: |f(x)| <= $x^2$ -> 0. Yes, continuous.

Differentiability at 0: f'(0) = lim [$h^2$ sin$\frac{1}{h}$]/h = lim h*sin$\frac{1}{h}$ = 0. Yes, differentiable.

But f' is discontinuous at 0: For x != 0: f'(x) = 2x sin$\frac{1}{x}$ - cos$\frac{1}{x}$. As x->0, the -cos$\frac{1}{x}$ term oscillates between -1 and 1. So lim f'(x) does not exist.

Lesson: f'(a) can exist even when f' is discontinuous at a. The derivative exists via the limit definition, but f' is not continuous.

Contrast with x*sin$\frac{1}{x}$, f(0)=0: f'(0) = lim sin$\frac{1}{h}$ = DNE. So f is continuous but NOT differentiable at 0.

Rule of thumb: $x^n$ * sin$\frac{1}{x}$: differentiable at 0 if n >= 2, not differentiable if n = 1.