Understanding f'(0) for f(x) = x^2sin(1/x) and Its Limit with Trig Functions

[endeavor]

Determine whether f'(0) exists.
f(x) = { x²sin (1/x) if x does not equal 0, 0 if x = 0}

$$f'(0) = \lim_{x\rightarrow 0} \frac{x^2 \sin \frac{1}{x}}{x - 0}$$
$$f'(0) = \lim_{x\rightarrow 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}}$$
Let u = 1/x. As x approaches 0, u approaches infinity. So I can't use lim x->0 (sin x)/x = 1. But now I have:
$$f'(0) = \lim_{u\rightarrow \infty} \frac{sin u}{u}$$
since sin u oscillates between -1 and 1, and u goes to infinity, f'(0) = 0.

 

[dimachka]

sure that is fine, you could also note that -x^2<=x^2*sin(1/x)<=x^2 and use that.