1. Sequences of Functions

I have been studying pointwise and uniform convergence of sequences of functions and i am stuck with this question,

Show that lim(Arctan nx) = (pi/2)sgnx for x in R.

I believe that we will have to take three cases for the limit function f(x), namely when it is equal to -(pi/2), 0 and (pi/2).
But i cannot figure out , a way to show that |fn(x) -f(x)|< epsilon.

Suppose you know that $\arctan(x)\to\pi/2$ as $x\to\infty$. That is, there is a function $\mu$ mapping neighborhoods of $\pi/2$ to neighborhoods of infinity: for every $\varepsilon>0$ and every $x>\mu(\varepsilon)$ we have $|\arctan(x)-\pi/2|<\varepsilon$.
Now fix an arbitrary x > 0. Let $N(\varepsilon)=\lceil\mu(\varepsilon)/x\rceil$ where $\lceil\cdot\rceil$ is the ceiling function. The essential property of $N(\varepsilon)$ is that $n>N(\varepsilon)$ implies $nx>\mu(\varepsilon)$. Therefore, for any $\varepsilon>0$ and $n>N(\varepsilon)$ we have $|\arctan(nx)-\pi/2|<\varepsilon$.
This (and a similar argument for x < 0) proves that $\arctan(nx)$ converges to $(\pi/2)\mathop{\text{sgn}}(x)$ pointwise. Note that it does not converge uniformly because $\sup_{x\in\mathbb{R}} |\arctan(nx)-\pi/2|= 1$ for all n.