# 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 $0$?

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.

$$g(x) = x$$ 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.