Critical point of a function Let $$f(x)=x^2\sin\left(\frac{1}{x}\right)$$
for $x\neq0$ and  $f(0) = 0$ .
Show that $0$ is a critical point of $f$ that is not a local maximum nor a local minimum nor an inflection point.
I need some help in order to show that.
If possible, give me only a hint, as it is homework and I only want some idea to finish the problem by myself. 
Thanks a lot . ; )
 A: *

*Prove it is a critical point (that the derivative exists and it is $0$).

*Prove it is neither a maximum or a minimum by proving that for any $\varepsilon>0$ there are $x$ and $y$, $0<x<\varepsilon,0<y<\varepsilon$  such as $f(x)>f(0)$ and $f(y)<f(0)$ (and $f(-x)<f(0)$ and $f(-y)>f(0)$) so there is no neighborhood in which $0$ is maximum or minimum.

*This also implies why it is not an inflection point.

A: Hint: think about what happens with the $\sin(1/x)$ part of the expression near zero. In a small neighborhood of the origin, you have infinitely many points of the form $2/k\pi$ (they accumulate near $0$) so your function oscillates infinitely many times between $-4/k^{2} \pi^{2}$ and $4/k^{2} \pi^{2}$ in that neighborhood, taking care of the maximum/minimum part of the question. Then look at the proper derivatives that give you the concavity and apply the same reasoning (notice that the $x^{2}$ factor is placed there exactly to give you sufficient smoothness to differentiate the function a certain number of times before it becomes discontinuous in $0$).
