Let $f_n(x)=\frac{1}{1+n^2x^2}$ for $n\in\mathbb{N},x\in\mathbb{R}$. Which of the following are true?
$f_n$ converges pointwise on $[0,1]$ to a continuous function
$f_n$ converges uniformly on $[0,1]$
$f_n$ converges uniformly on $[\frac{1}{2},1]$
$\lim_{x\to\infty}\int_{0}^{1}f_n(x) dx=\int_{0}^{1}\left(\lim_{x\to\infty}{f_n(x)}\right)dx$
My work. 1) if $x=0$, $f_n(x)=1$ which converges to $1$ as $n\to{\infty}$,
if $x=1$, $f_n(x)=\frac{1}{1+n^2}$ which converges to $0$ as $n\to{\infty}$,
if $x\in\left(0,1\right)$ then $f_n$converges to 0 as $n\to{\infty}$.
Hence $f_n$ converges pointwise on $[0,1]$ to a function which is not continuous. Hence option 1. 2. and 4. are wrong.
3) Consider $\vert{f_n(x)-f(x)}\vert=\vert{\frac{1}{1+n^2x^2}-0}\vert$ which converges to $0$ as $n\to{\infty}$ for all $x\in[\frac{1}{2},1]$. Hence it is uniformly continuous on $[\frac{1}{2},1]$.
Is my justification correct?
You have to revise partially your work. Your answers to 1) and 2) are correct. As regards 4), note that as $n\to\infty$, $$\int_{0}^{1}\frac{1}{1+n^2x^2}\, dx=\left[\frac{\arctan(nx)}{n}\right]_0^1=\frac{\arctan(n)}{n}\to 0.$$ What is the integral of the poitwise limit $f$?
For 3), you may say that as $n\to\infty$, $$\sup_{x\in [\frac{1}{2},1]}\vert{f_n(x)-f(x)}\vert=\sup_{x\in [\frac{1}{2},1]}\left|\frac{1}{1+n^2x^2}-0\right|=\frac{1}{1+\frac{n^2}{4}}\to 0$$ (because $f_n(x)$ is decreasing and positive in $[0,1]$) and therefore the sequence converges uniformly in $ [\frac{1}{2},1]$.