I never took a class where uniform convergence was introduced, so I only know the Definition and not much more about it.
I have an Approximation of some sequence of functions $f_n$ by some function $g$. I know that if we introduce $\varepsilon(x)$ as the relative error, thus $f_n(x)= g(x)\left(1+\varepsilon_n(x)\right)$, that $\varepsilon_n(x) < \varepsilon_n(y)$ for $x >y$, thus the error decreases with increasing function Argument. We further know that $\varepsilon_n(x) \rightarrow 0$ for every fixed $x=O(1)$.
Is this enough to show that the convergence of $f$ to $g$ is uniform in $x$?
NOTE: This only answers the first version of the question.
To answer your question, on $(0,+\infty)$, let $f_n(x) = x + \frac{1}{n}\sqrt{x}$, and $g(x) = x$. Then the relative error $1/n\sqrt{x}$ decreases with $x$ and tends to zero for any fixed $x$. However, the absolute error is $\sqrt{x}/n$, which is unbounded for all $n$. Thus $f_n$ cannot converge uniformly to $g$.