I am trying to show the following,
Let $f : \mathbb{R^+} \to \mathbb{R}$ be a real-valued function such that $$\lim_{x \to \infty} f(x) = \lim_{x \to 0} f(x) = 0$$
Let $g : \mathbb{R^+} \to \mathbb{R}$ be a real-valued function such that $$g(x) = f(x)\left[1+\text{sin}\left(2 \pi \text{ln}(x)\right)\right]$$
Show that $\lim_{x \to \infty} g(x) = \lim_{x \to 0} g(x) = 0$.
My attempt:
We know $g(x) = f(x)\left[1+\text{sin}\left(2 \pi \text{ln}(x)\right)\right]$ and $$ 0 \leq \left[1+\text{sin}\left(2 \pi \text{ln}(x)\right)\right] \leq 2$$
Hence, $$\lim_{x \to \infty} g(x) = \lim_{x \to \infty} f(x) = 0$$ and $$\lim_{x \to 0} g(x) = \lim_{x \to 0} f(x) = 0$$
Is this correct / rigorous enough? Is there a better way of doing this? I am not sure that the fact that $1 + \text{sin}(2 \pi \text{ln}(x))$ is bounded is enough, since some limits don't converge due to rapid oscillation. How can I make my attempt better?
Does this principle apply in general when we have some limiting behaviour of a function $f$, and a bounded function $g$, is the limiting behaviour of $fg$ the same as that of $f$?
You should clarify the use of the squeeze theorem: since $0\le 1+\sin(2\pi\ln x)\le 2$, we also have $$ 0\le g(x)\le 2f(x) $$ and since $\lim_{x\to?}f(x)=0$ (where $?=\infty$ or $?=0^+$), we can conclude that $$ \lim_{x\to?}g(x)=0 $$
This uses substantially that the limits of $f$ are both zero, otherwise the application of the squeeze theorem cannot be done. A simple example: $f(x)=1$ and $g(x)=f(x)(1+\sin(2\pi\ln x))$. Neither the limit at $0^+$ nor the limit at $\infty$ exist.