Determine the limit, or show it doesn't exist: $\lim_{x\to 2} \left(\arctan\left(\frac{1}{2-x}\right)\right)^2$ 
Determine the limit of the following or prove it doesn't exist: 
  $$\lim_{x\to 2} \left(\arctan\left(\frac{1}{2-x}\right)\right)^2$$

If I just plug in the value of $x$, I get an undefined expression. But, unfortunately, I don't see how to expand this limit to either see it doesn't exist or get a value. Any help would be much appreciated.
 A: Performing the substitution $u=\frac{1}{2-x}$, this is just $\lim\limits_{u\to\infty}(\tan^{-1} u)^2$, which evaluates to $\frac{\pi^2}{4}$, since $\lim\limits_{u\to\infty}\tan^{-1} u=\frac{\pi}{2}$. (Note that there actually is a slight technicality, namely that we took only the right hand limit of the original integral. Luckily, the fact that the arctan is squared makes the left hand limit consistent.)
A: Intuitively, as $x$ 2 from the positive side, $\frac{1}{2-x}$ approaches $-\infty$ and as $x$ approaches 2 from the negative side, $\frac{1}{2-x}$ approaches $\infty$.  However, $arctan$ of $\pm\infty$ is $\pm\frac{\pi}{2}$, so squaring this gets $\frac{\pi^2}{4}$, the value of the limit. 
Just because, let's do it rigorously.
We show that the left and right limits exist and that they are equal, taking for granted that $\lim_{x\to\infty}arctan(x)=\frac{\pi}{2}$ and that $\lim_{x\to -\infty}=-\frac{\pi}{2}$ and that $arctan(x)$ is monotonic.
Consider $\lim_{x\to2^{-}}arctan^2(x)$.  Pick $Y$ s.t. $y>Y\rightarrow \frac{\pi}{2}-arctan(y)<\epsilon$ so then $\frac{\pi^2}{4}-arctan(y)<\epsilon(\pi-\epsilon)$.  To get $y>Y$, we need $\frac{1}{2-x}>Y\rightarrow x>2-\frac{1}{Y}$ and, since this is the left limit, $x<2$.  So take $\delta=2-\frac{1}{Y}$ so that $x\in(2-\delta,2)\rightarrow arctan^2(\frac{1}{2-x})\in\Bigl(\frac{\pi^2}{4}-\epsilon(\pi-\epsilon),\frac{\pi^2}{4}\Bigr)$, so $\lim_{x\to2^-}arctan^2(x)=\frac{\pi^2}{4}$
Excercise: compute $\lim_{x\to2^+}arctan^2(x)$ and show they are equal.
