That sounds a lot like a tongue-twister. I know that there exist sequences of Lipschitz functions whose uniform limit is not Lipschitz (for instance, just use Weierstrass theorem on $[a,b]$). Clearly if the sequence is uniformly Lipschitz, then the uniform limit has to be Lipschitz.
I think something similar should happen when we replace Lipschitz with uniform continuity. Weierstrass-Stone theorem though cannot be used as it requires a compact interval of definition which, in turn grants uniform continuity of continuous functions.
Is there an easy counterexample or is it actually true that
Given $\{f_n\}$ a sequence of uniform continuous functions, let $f$ be such that
$$
f_n \to f
$$
uniformly. Is then $f$ uniformly continuous itself?
Best Answer
$f$ must be uniformly continuous.
Proof: Let's choose any $\epsilon\gt 0$.
Because of uniform convergence, there is one of the functions (say $f_n$) such that $|f_n(x)-f(x)|\lt\frac{\epsilon}{3}$ for all $x$.
Now, $f_n$ is uniformly continuous, so let's find $\delta\gt 0$ such that $|x-y|\lt\delta$ implies $|f_n(x)-f_n(y)|\lt\frac{\epsilon}{3}$.
Finally, for any $x,y$ such that $|x-y|\lt\delta$, we have $|f(x)-f(y)|\le|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|\lt\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$
As $\epsilon\gt 0$ was arbitrarily chosen to start with, it follows that $f$ is uniformly continuous.