# 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.

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.)

• Could you please elaborate on how you arrived at that conclusion, i'm not sure i'm following. Commented Dec 18, 2018 at 3:06
• OK I added a bit Commented Dec 18, 2018 at 3:10

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.