A uniform limit of real-analytic functions certainly need not be real-analytic. Any continuous function on $[a,b]$ is a uniform limit of polynomials (and is hence the sum of a uniformly and absolutely convergent series of polynomials).
I can't think of any "simple" criteria. Even uniform convergence of a sequence of functions together with uniform convergence of every derivative is not enough; consider the Poisson integral of a function in $C^\infty_c$.
METHODOLOGY $1$:
Note that for $x=1-1/n$, $x^n\to e^{-1}$.
Hence, $\displaystyle \sup_{x\in [0,1)} x^n\ne 0$ and the sequence does not converge to $0$ uniformly.
METHODOLOGY $2$:
Alternatively, take $\epsilon=1/4$. Then, for all $N\ge 1$ there exists a number $x=1-1/n$ and a number $n>N$ such that
$$x^n=\left(1-\frac1n\right)^n\ge \epsilon=\frac14$$
since $\left(1-\frac1n\right)^n$ increases monotonically.
Therefore, from the definition of the negation of uniform convergence we conclude that $x^n$ does not uniformly to $0$.
SHOWING UNFORM CONVERGENCE FOR $\displaystyle x\in [0,r], 0<r<1$
We can show that $x^n$ converges uniformly on any compact subset of $[0,1)$. Take $x\in [0,r]$ for $0<r<1$. Then, we have for all $\epsilon>0$
$$x^n\le r^n<\epsilon$$
whenever $n>\max\left(1,\log(\epsilon)/\log(r)\right)$.
Best Answer
Choose your favorite continuous, non-real-analytic function on a compact interval, and then apply Weierstrass' approximation theorem to it. Bernstein polynomials can give a constructive example if you want.