If $f$ is uniformly continuous, how can I show that $g$ is uniformly continuous

Question

1. $f,g:\mathbb{R}\rightarrow \mathbb{R}$ are continuous functions
2. $\forall x \in \mathbb{R}\quad ;\quad g(x)\neq 0\quad$; $\quad g$ is bounded
3. $$\lim_{x \to +\infty}{\frac{f(x)}{g(x)}} =1$$

If $f$ is uniformly continuous, how can I show that $g$ is uniformly continuous too?

Someone has given a wrong counterexample to show that the question is wrong by comparing $-x$ and $e^{-x}$. Why is this counterexample wrong to show my question is wrong?Because I have already said that condition $3$ stated in question should hold for it to be true.The counterexample stated does not fulfill the $3$rd condition since it is wrong hence does not prove anything.

Someone else has claimed the question is false without giving proper reason for why my question should provide information about "what happens to $\frac{f}{g}$ as $x\rightarrow -\infty$"

What I have tried? $$\forall\epsilon\hspace{3mm}\exists c\hspace{3mm} \forall x:\hspace{3mm}\left\| \frac{f(x)}{g(x)}-1 \right\|<\epsilon$$ $$\implies \hspace{3mm}\left\| \frac{f(x)-g(x)}{g(x)} \right\|<\epsilon$$ $$\epsilon = \frac{1}{2}\quad; \quad \frac{g(x)}{2}<f(x)<\frac{3}{2}g(x)$$
I have no idea on how to use second condition $2$ in favor of solving the problem.
The working example would be consider both $f$ and $g$ functions to be $$f(x)=\frac{1}{1+x^2}$$ $$g(x)=\frac{1}{1+x^2}$$ I could not find an example in which $f$ and $g$ would be different functions that fulfills the question conditions.

Answer 1

HINT: $|\frac12f(x)|\leq|g(x)|\leq\frac32|f(x)|$ for large $x$.

Some additional assumptions about behaviour at $-\infty$ are needed, e.g. $\lim_{x\to\pm\infty}\frac{f(x)}{g(x)}=1$, because without it the thesis is not true: compare $-x$ and $e^{-x}$.

Answer 2

You forget to say what happens to $f/g$ as $x \to -\infty$ so the assertion is false.