The monotonicity of function $y=x/2 + x^{2}\sin(1/x)$ near $x=0$ Is the function $y=x/2 + x^{2}\sin(1/x)$ monotonic near $0$? 
The derivative $f'$ obviously goes positive and negative near $0$, because $$f'(x)= \frac12 + 2x\sin(1/x) - \cos(1/x))$$ Does that mean that $f$ is not monotonic near $0$? 
 A: It may be appropriate to now summarize what is spread across comments into an answer.
The function 
$$f(x) = \begin{cases}\tfrac x2+x^2\sin(1/x)&\text{if }x\ne 0\\
0&\text{if }x=0\end{cases}$$
has one property that may make it look like being monotonic near $0$: 
For $0<|x|<\frac 12$ we have $|x^2\sin(1/x)|\le|x^2|<\left|\frac x2\right|$, hence $f(x)>0$ for $0<x<\frac12$ and $f(x)<0$ for $-\frac12 < x < 0$, just as would be expected from a function monotonic near $0$ (the result may be called monotonic at $0$, but I'm not sure if that concept is in use).
However, we say that $f$ is monotonic near $0$ if there exists an open neighbourhood $U$ of $0$ such that the restriction $f|_U$ of the given function $f$ to that neighbourhood $U$ is monotonic.
As you already noted,  for $x\ne0$ we have 
$$f'(x)= \frac 12 + 2x\sin(1/x) - \cos(1/x).$$
Let $U$ be an open neighbourhood of $0$.
For $n\in \mathbb N$ big enough, we have $\xi:=\frac1{2n\pi}\in U$.
Then we check that $f'(\xi)=-\frac12<0$.
Since $U$ is open, we have $\xi+h\in U$ for all sufficiently small positive $h$.
By the definition of derivative, for all sufficiently small positive $h$ we have that $\left|\frac{f(\xi+h)-f(\xi)}{h}-f'(\xi)\right|<\frac12$ and hence $f(\xi+h)<f(\xi)$.
On the other hand, we already saw that $f(\xi)>0$ because $0<\xi<\frac12$.
Then from $f(0)<f(\xi)>f(\xi+h)$ we see that $f|_U$ is not monotonic.
