Does the sequence $f_n(x)={1+x^n\over 2+x^n}$ converge uniformly on $[0,1)$? Does the sequence of functions $$f_n(x)={1+x^n\over 2+x^n}$$ 
converge uniformly to ${1\over 2}$ on the interval $0\le x\lt1$?
I don't think so because  $\sup\left|f_n(x)-{1\over2}\right|=1$. If the interval is $[0,a]$ where $0\lt a\lt1$, then I think this is true. 

You're right, it is not uniformly convergent. And given $\varepsilon>0$, you can even find how badly $N(x)$ degenerates as $x$ approaches 1, where $N(x)\in\mathbb{N}$ is such that $|f_n(x)-0.5|<\varepsilon,\forall n>N(x)$. Notice that
$$
f_n(x) = 1 - \frac{1}{2+x^n}
$$
so
$$
\left|f_n(x)-\frac{1}{2}\right| = \frac{x^n}{4+2x^n}<\varepsilon
$$
If you solve for $x$ you obtain $x^n<\frac{4\varepsilon}{1-2\varepsilon}$, which gives (recall that $x<1$, so $\log_x(b)$ is decreasing)
$$
n>\log_x\frac{4\varepsilon}{1-2\varepsilon}=\frac{\ln\frac{4\varepsilon}{1-2\varepsilon}}{\ln x}
$$
So $N(x)$ blows up roughly as $1/(x-1)$ as $x$ approaches 1.
Edit: yes, if you restrict your attention to $[0,a]$ with $0<a<1$, then $N(x)$ as a maximum, so the sequence uniformly converges.
