why $\arctan(nx)$ doesn't converge uniformly? 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!
 A: Set $f_n(x)=\arctan(nx)$. Then  $\{f_n\}$ converges pointwise to $f(x)={\pi\over2}$ on $(0,\infty)$.
If $x_n={1\over  n}$, then $f_n(x_n)=\arctan(1)={\pi\over4}$. So, the sequence $\{f_n\}$ does not converge uniformly to $f$ on $(0,\infty)$. 
You have to compute the $\sup$ for each fixed value of $n$ first, and then take the limit.
From the above, we have  for any $n$:$$\sup\limits_{x\in(0,\infty)}|f_n(x)-f(x)|\ge {\pi\over 2}-{\pi\over 4}={\pi\over 4 }.$$   From this, $$\lim\limits_{n\rightarrow\infty} \sup\limits_{x\in(0,\infty)}|f_n(x)-f(x)|\ne 0.$$ 
 A picture may help:


Note that, in fact,  $\sup\limits_{x\in(0,\infty)}|f_n(x)-f(x)|={\pi\over2}$ for all $n$. 
A: Uniform convergence means that the distance between $f_n(x)$ and $f(x)$ becomes small for all $x$, at the same time. To emphasisze here are the two definitions:


*

*$f_n$ converges to $f$ pointwise if for every $x$ and every $\varepsilon >0$ there exists $n_0$ such that $|f(x)-f_n(x)|<\varepsilon$ for all $n \geq n_0$.

*$f_n$ converges uniformly to $f$ if for every $\varepsilon >0$ there exists $n_0$ such that $|f(x)-f_n(x)|<\varepsilon$ for all $n \geq n_0$ and for all $x$.
In your case, the problem is that every function $f_n$ can be extended continuously by $f_n(0)=0$, so near $0$ the functions $f_n$ cannot approach $\pi/2$ simultaneously for small $x$. If your functions would have been defined on $(\delta,\infty)$ with $\delta>0$ then the convergence would have been uniform.
Another way to see that $f_n$ does not converge uniformly to $f$ is to consider $f_n :[0,\infty)$, extended by continuity. It is relatively easy to prove that if $f_n$ converges uniformly to $f$ and all $f_n$ are continuous then $f$ is also continuous. But the limit of your functions $f_n(x)=\arctan (nx)$ is
$$ f(x)=\begin{cases}\frac{\pi}{2},& x>0 \\ 0,& x=0\end{cases}$$
which is not continuous.
A: Here is another reason to see that the convergence is not uniform.  The functions $\arctan(nx)$ converge pointwise to the function that is 0 at 1 and $\pi/2$ for any positive number.  Each function is continuous, but the limit  is not.  Hence, the convergence cannot be uniform.
