$f(x) = \begin{cases}x + 2x^2\sin\left(\frac1x\right),& x\ne 0\\0,& x = 0\;.\end{cases}$
I am having a tough time answering this question in a rigorous mathematical way, here is what I have tried:
I have proved in a previous part of this question that $f\,'(0) = 1$.
The derivative when $x\ne 0$ is,
$f\,'(x) = 1+4x\sin\left(\frac1x\right)−2\cos\left(\frac1x\right)\;.$
I have $\lim\limits_{x\to 0} f\,'(x) = 1 - 2\cos\left(\frac1x\right)\;,$ which oscillates between $3$ & $-1$.
Since $f\,'(x)$ containing $0$, is not continuous, it cannot be increasing on the interval. Am I on the right track here?
Thanks in advance!