Does $\lim_{x\to 0} \frac{f(x)}{g(x)}=1$ and that $g(x)$ is increasing near $0$ imply that function $f(x)$ is also increasing near $0$?

Question

Suppose $f(x)$ and $g(x)$ are two real functions and both $\to 0$ as $x\to 0$. My claim: Given that $\lim_{x\to 0} \frac{f(x)}{g(x)}=1$ and $g$ is non-decreasing on the interval $(0,\delta)$ for some $\delta >0$, the function $f(x)$ is also non-decreasing on the interval $(0,\delta)$?

One example is $f(x)=\sin(x)$ and $g(x)=x$. More generally, by l'hopital's theorem,

$$ \lim_{x\to 0} \frac{f'(x)}{g'(x)} =\lim_{x\to 0} \frac{f(x)}{g(x)} =1.$$ So my claim is true for all $g$ and $f$ differential on a right neiborhood of $0$. But this gives no clue whether my claim is true for all real function with the given conditions. I also tried to find a function $f$ that oscillates a lot and a function $g$ monotone near $0$ such that $\lim_{x\to 0} \frac{f(x)}{g(x)} =1$. They would be a couterexample if exist. But I failed finding them.

Answer 1

$g(x) = x$ and $f(x)=x + 4x^2 \sin\left(\frac{1}{x}\right)$ is a counterexample to your claim.

See here for proof that $f$ is not non-decreasing in any neighborhood of zero.