Calculus Proof Unsure Let $$g(x)=x^2\sin\left(\frac{1}{x}\right)+\frac{1}{2}x$$ Show that $g'(0)>0$ but there is no neighborhood of $0$ on which $g$ is increasing. (More precisely, every interval containing $0$ has sub intervals on which g is decreasing). 
For the first part, I used the limit definition of the derivative to calculate $g'(0)=0.5$. I have an intuitive understanding of the second part but I am having trouble coming up with an approach as to how to come up with the proof for it. Any help would be appreciated.

The idea is that if there were a neighborhood of $0$ for which $g$ is increasing, then there would exist some $\epsilon > 0$ such that for every $x \in (-\epsilon, \epsilon)$, $g'(x) > 0$.  But this is not the case if we were to calculate the critical points of $g$.
