Suppose $y:[0, \infty)\to [0,\infty)$ is differentiable with $y(0)=0$ and such that for every $\delta>0$ we have that $y|_{0,\delta}\not \equiv 0$.
Does it follow that there exists $\varepsilon>0$ is strictly monotonic over $[0,\varepsilon)$?
This seems intuitively true, but I was not able to prove it.
On
The answer is no.
In fact for every positive $\epsilon$ we can find a counter example.
Note that for a given $\epsilon \ge 0$ the function $$f(x)=sin( \frac {3\pi x}{4 \epsilon })$$ is a continuous function on $(0,\epsilon)$ and $ \lim_{x\rightarrow 0^{+}}f(x)=0$ while the function is not monotonic on$(0,\epsilon)$ because it has a local maximum at $x= \frac {2\epsilon}{3}$.
Answer for the before edited question:
Take $f(x)=|\sin x|$ and $\epsilon=\pi$, then $f(x)=\sin x\ne 0$ on $(0,\pi)$, and $\lim_{x\rightarrow 0^{+}}f(x)=\lim_{x\rightarrow 0^{+}}\sin x=0$, but $f$ is not strictly monotonic on $(0,\pi)$?
Answer for the edited question:
Take $f(x)=x(\sin(1/x)+2)$ for $x>0$ and $f(0)=0$. Then $f(x)>0$ for $x>0$. And $f'(x)=\sin(1/x)+2-(1/x)\cos(1/x)$ for $x>0$. For large $n$, $\dfrac{1}{2n\pi+\dfrac{\pi}{4}}\leq x\leq\dfrac{1}{2n\pi}$, $f'(x)<0$ for such an $x$. For large $n$, $\dfrac{1}{2n\pi+\dfrac{3}{4}\pi}\leq x\leq\dfrac{1}{2n\pi+\pi}$, $f'(x)>0$ for such an $x$, so $f$ is not monotonic on each $(0,\epsilon)$.