$(f_n(x))_{n=1}^{\infty}$ , $g \in C[\mathbb{R}]$. Prove: $(g\circ f_n)_{n=1}^{\infty}$ uniformly converges iff $g$ is uniformly continues Let $(f_n(x))_{n=1}^{\infty}$ a series of uniformly continuous functions, $\mathbb{R}\to\mathbb{R}$ which uniformly converges to the function $f$, and a continuous function $g$ :$\mathbb{R}\to\mathbb{R}$.
I need to give an example where $(g\circ f_n)_{n=1}^{\infty}$ is not uniformly converges and to prove that if $g$ would be uniformly continuous so $(g\circ f_n)_{n=1}^{\infty}$ would uniformly converges.
As an example I gave $g=\sin x$ and $f_n= \frac{1} {n+x^2}$, but I'm having a hard time proving the claim.
Any hints? Thanks!
 A: To prove the second claim, now assume that $g$ is uniformly continuous. By definition of uniformly continuous, for all $\epsilon>0$, there exists $\delta>0$ which depend only on $\epsilon$ such that 
$$|g(x)-g(y)|<\epsilon\mbox{ for all $x, y\in\mathbb{R}$ such that }|x-y|<\delta.$$
Since $f_n$ converges uniformly to $f$, for the above $\delta>0$, there exists an positive integer $N$ such that if $n\geq N$, then
$$|f_n(x)-f(x)|<\delta\mbox{ for all }x\in\mathbb{R},$$
which implies that  if $n\geq N$, then 
$$|g\circ f_n(x)-g\circ f(x)|=|g(f_n(x))-g(f(x))|<\epsilon\mbox{ for all }x\in\mathbb{R}.$$
This proves the claim.
A: For the counterexample:
Take $f_n(x)=x+{1\over n}$, $f(x)=x$, and $g(x)=x^2$. Then each $f_n$ is uniformly continuous, and $\{f_n\}$ converges uniformly to $f$.
Now, $g(f_n(x))=\left(x+{1\over n}\right)^2 $ converges pointwise to $h(x)=x^2$. 
We have:
$$
|(x+{\textstyle{1\over n}})^2 -x^2|= \left|{2 x \over n} +{1\over n^2}\right|,
$$
which can be made arbitrarily large for any fixed $n$. This shows $g\circ f_n$ does not converge uniformly to $h$.
