I'd love your help with understanding what led me to conclude that $\arctan(nx)$ converges uniformly for $x \in (0, \infty)$
I wanted to check that $\lim_{n \to \infty} \sup |f_n(x)-f(x)|$ is 0 where $f(x)=\lim_{n \to \infty} f_n (x)$.
$f(x)=\frac {\pi}{2}$.
I interpret $\lim_{n \to \infty} \sup |f_n(x)-f(x)|$ as: for every $x_0$ in my interval, compute $\lim_{n\to \infty} |f_n(x_0)-f(x)|$=$L_{x_0}$ and then take the $\sup (L_{x_0})$ for all $x_0$ in my range. Am I right?
So, In my case I have $f(x)=\lim \arctan (nx)= \frac {\pi}{2}$
$r_n(x)=f_n(x)-f(x)=\frac {\pi}{2}-\arctan(nx)$
So for every $x \in (0, \infty)$ , small as can be, when I'll compute $\lim \frac {\pi}{2}-\arctan(nx) =0$ when $n \to \infty$ so $\sup (L_{x_0})=0$ since all the values in this set is $0$. What do I get wrong here? what is the correct defenition for$\lim_{n \to \infty} \sup |f_n(x)-f(x)|$? I feel like I got it all wrong.
Thanks guys!