Why doesn't the sequence of functions $f_n(x)=\arctan{nx}$ converge uniformly?

Question

I have a sequence of functions

$f_n(x)=\arctan{nx}$

The function to which this series converges is

$f(x):=\lim_nf_n(x)$

My book tells me that the sequence doesn't converge uniformly since

$\sup_{x>0}|f(x)-f_n(x)|=\frac{\pi}{2}\nrightarrow0$

Why does that equal $\frac{\pi}{2}$?
I must be getting something wrong, but I rewrote the formula as:

$\sup_{x>0}|\frac{\pi}{2}-\arctan{nx}|$

Which should give $0$, so I must've made a mistake somewhere...

Thanks in advance!

Answer 1

Note that $\arctan nx$ is an increasing function so $\frac{\pi}{2} - \arctan nx$ is a decreasing function. As $\arctan 0 = 0$ and $\arctan nx$ is bounded above by $\frac{\pi}{2}$, the function $\frac{\pi}{2} - \arctan nx$ is positive for $x > 0$. Therefore $$\sup_{x > 0}\left|\frac{\pi}{2}-\arctan nx\right| = \sup_{x > 0}\left(\frac{\pi}{2}-\arctan nx\right) = \lim_{x\to 0^+}\left(\frac{\pi}{2}-\arctan nx\right) = \frac{\pi}{2}.$$