An example of a locally uniformly convergent sequence which does not converge uniformly I'm looking for an example of a sequence of real-value functions $\left\{ f_{n}:\mathbb{R}\to\mathbb{R}\ |\ n\geq1\right\}$ 
  which converges locally uniformly to some function $f$ on some interval $\left(a,b\right)\subseteq\mathbb{R}$
  but does not converge uniformly there. Just for clarity, the definitions involved are:
A sequence of functions $f_{n}$
  is said to converge uniformly to a function $f$
  on $\left(a,b\right)$
  if $$\sup\limits _{x\in\left(a,b\right)}\left|f_{n}\left(x\right)-f\left(x\right)\right|\overset{n\to\infty}{\longrightarrow}0$$
The sequence is said to converge locally uniformly to $f$
  on $\left(a,b\right)$
  if for each $x\in\left(a,b\right)$
  there is a neighborhood $U$
  of $x$
  such that $f_{n}$
  converges uniformly to $f$
  on $\left(a,b\right)\cap U$
 .
I did some searching and I found a couple of places that suggest looking at $f_{n}\left(x\right)=x^{n}$
  on $\left(0,1\right)$
  but it seems to me this sequence simply converges uniformly to $f\equiv0$
  on the entire open interval.
 A: Consider the following sequence of functions $f_n:\mathbb{R}\to\mathbb{R}$.$$f_n(x)=\left\{\begin{array}{rc}-n&x<n\\x&-n\leq x\leq n\\n&n<x\end{array}\right..$$This sequence converges uniformly locally to $f(x)=x$, but does not converge uniformly to any function.
Apply any homeomorphism $\mathbb{R}\to(0,1)$, and get your desired sequence.
A: $f_n(x)=x^n$ does not converge uniformly to zero on $(0,1)$. It does converge pointwise to zero on the entire interval, but the convergence is very slow near $1$. 
More precisely, to have $x^n<\varepsilon$ it is equivalent to have $n \ln(x)<\ln(\varepsilon)$ so $n>\frac{\ln(\varepsilon)}{\ln(x)}$. As $x \to 1^-$ the denominator goes to zero, so you do not have globally uniform convergence. On the other hand, on compact subsets $\ln(x)$ is bounded away from zero, and so you have locally uniform convergence.
Note that if it did converge uniformly, then we would have 
$$\lim_{n \to \infty} \lim_{x \to 1^-} x^n = \lim_{x \to 1^-} \lim_{n \to \infty} x^n.$$
But that equality reads $1=0$.
