Prove that $f(x) = \frac{x}{2} + x^2\sin(\frac{1}{x})$ is monotone near $x=0$ Let $f(x) = \frac{x}{2} + x^2\sin(\frac{1}{x})$.
I need to prove that there is a neighborhood of 0 that f is monotone there.
I tried to check the sign of the first derivative and then conclude something about monotonicity (using Lagrange mean value theorem), but couldn't really make a progress.
I'll be happy for help from you guys.
Thanks. 
 A: It's not true:
First, we need to define $f(0)=0$. Then we  have:
$$\tag{1}f'(x)={1\over 2} +2x\sin(1/x)+ \cos(1/x)\quad\text{for}\quad x\ne 0.$$
and
$$f'(0)
=\lim\limits_{h\rightarrow 0} { (h/2)+h^2\sin(1/ h)\over h } 
=\lim\limits_{h\rightarrow 0}\bigl[\, { {\textstyle{1\over2}}+h\sin(1/ h)  }\,\Bigr]={\textstyle{1\over 2}}.$$
So, $f$ is differentiable everywhere.
To show that $f$ is monotone on no interval, use the
Hint: For $x$ near $0$, what can you say about the sign of $f'$?


Solution:
In any neighborhood $O=(-\delta,\delta)$ of $0$:
For any $x\in O$ the term $2x\sin(1/x)$ of $(1)$ is small. 
In fact, we have  $|2x\sin(1/x)|\le 2|x|$.  But we can find values  $x_1$ and $x_2$ in $O$ as small as we like such that $\cos(1/x_1)=-1$ and   $\cos(1/x_2)= 1$.
Keeping these thoughts in mind and looking at the form of $f'$ given by  $(1)$,  it follows that 
$f'$ takes both positive and negative values in $O$. 
Thus, since a differentiable function is monotonic on an interval if and only if its derivative does not change sign on the interval, $f$ is monotonic in no neighborhood of 0.
