$f_n(x)= \frac{1+x^n}{2+x^n},$ $x\in[0,1)$
Clearly, the pointwise limit function is the constant $\frac{1}{2}$ function.
I found $|f_n(x)-f(x)|$ for uniform convergence:
Let $g_n(x)=|f_n(x)-f(x)|$.
$g_n(x)=\frac{x^n}{4+2x^n}$.
I noticed that $g_n((\frac{1}{2})^\frac{1}{n})= \frac{1}{10} \neq 0$.
Therefore we don't have uniform convergence in the given interval. I have here 2 questions. Does my work seem good and acceptable? Also, what if we want to change the domain of definition to $[a,1)$ or $[0,a]$ for some $a$? How should we choose this $a$ to avoid having $x's$ like the one I mentioned before in the interval?
Regards.
The problem (for the uniform convergence) is not near $x=0$, but near $x=1$. (Indeed the sequence $(1/2)^{1/n}$ converges to $1$.)
You can see this fact also in this way. If you consider the functions $f_n$ defined in $[0,1]$, instead of $[0,1)$, then the sequence converges pointwise to the function $$ f(x) = \begin{cases} 1/2, &\text{if}\ x\in[0,1),\\ 2/3, &\text{if}\ x = 1, \end{cases} $$ hence $(f_n)$ cannot converge uniformly to $f$ in $[0,1]$ (for otherwise $f$ must be continuous, being the $f_n$ continuous). But you see in a moment that $$ \sup_{x\in[0,1]} |f_n(x) - f(x)| = \sup_{x\in[0,1)} |f_n(x) - f(x)|, $$ so the convergence cannot be uniform also in $[0,1)$.