Find the value of the given limit. 
The value of $\lim_{x\to \infty} (x+2) \tan^{-1} (x+2) - x\tan^{-1} x $ is $\dots$
a) $\frac{\pi}{2} $ $\qquad \qquad \qquad$ b) Doesn't exist $\qquad \qquad \qquad$ c) $\frac{\pi}{4}$ $\qquad \qquad$ d)None of the above.

Now, this is an objective question and thus, I expect that there must be an easier way to do it either by analyzing the options or something else. I'm not sure about this though!
What I've done yet is trying to apply : $\tan^{-1}a - \tan^{-1}b$ formula but it requires certain conditions to hold true which are not specified here. I'm not sure how we  will approach this! Any kind of help will be appreciated.
 A: Let $a=\tan^{-1}(x+2),b=\tan^{-1}x$.
Then, since one has $\tan a=x+2,\tan b=x$, one has$$\begin{align}\lim_{x\to \infty}a\tan a-b\tan b&=\lim_{x\to\infty}a(\tan b+2)-b\tan b\\&=\lim_{x\to\infty}2a+(a-b)\tan b\\&=\lim_{x\to \infty}2\tan^{-1}(x+2)+\frac{\tan^{-1}(x+2)-\tan^{-1}x}{\frac 1x}\\&=2\cdot\frac{\pi}{2}+\lim_{x\to\infty}\frac{\frac{1}{1+(x+2)^2}-\frac{1}{1+x^2}}{-\frac{1}{x^2}}\\&=\pi+\lim_{x\to\infty}\left(-\frac{x^2}{1+(x+2)^2}+\frac{x^2}{1+x^2}\right)\\&=\pi-1+1\\&=\pi\end{align}$$
A: you can use the mean value theorem on 
$$f(x) = x\tan^{-1} x, \quad f'(x)= \tan^{-1} x + \frac{x}{1+x^2} \to \pi/2 \text{ as } x \to \infty.$$
by mvt, $$(x+2)f(x+2) - xf(x) = 2f'(x+\epsilon) \to \pi \text{ as } x \to \infty.$$
A: Since
$$
\tan^{-1}(x)=\frac\pi2-\tan^{-1}\left(\frac1x\right)=\frac\pi2-\frac1x+O\left(\frac1{x^3}\right)
$$
we get
$$
x\tan^{-1}(x)=\frac\pi2x-1+O\left(\frac1{x^2}\right)
$$
Thus,
$$
(x+2)\tan^{-1}(x+2)-x\tan^{-1}(x)=\frac\pi2\cdot2+O\left(\frac1{x^2}\right)
$$
so that
$$
\lim_{x\to\infty}\left[(x+2)\tan^{-1}(x+2)-x\tan^{-1}(x)\right]=\pi
$$
A: Here again ! When $y$ is large, an asymtotic expansion is $$\tan^{-1}(y)=\frac{\pi }{2}-\frac{1}{y}+\frac{1}{3 y^3}+O\left(\left(\frac{1}{y}\right)^4\right)$$ $$A=(x+2) \tan^{-1} (x+2) - x\tan^{-1}(x)$$ $$A\approx(x+2)\Big(\frac{\pi }{2}-\frac{1}{x+2}+\frac{1}{3 (x+2)^3}\Big)-x\Big(\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}\Big)=\pi -\frac{4 (x+1)}{3 x^2 (x+2)^2}$$ where you see the limit and how it is approached.
Again, plot the function and the approximation for $0 \leq x\leq 10$ and tell me if this is funny or not.
A: Set $1/x=h$ to get $$\lim_{h\to0^+}\dfrac{(1+2h)\tan^{-1}\dfrac{1+2h}h-\tan^{-1}\dfrac1h}h$$
$$=\lim_{h\to0^+}\dfrac{\tan^{-1}\dfrac{1+2h}h-\tan^{-1}\dfrac1h}h+2\lim_{h\to0^+}\tan^{-1}\dfrac{1+2h}h$$
Now,
$$\tan^{-1}\dfrac{1+2h}h-\tan^{-1}\dfrac1h=\tan^{-1}\left[\dfrac{\dfrac{1+2h}h-\dfrac1h}{1+\dfrac{1+2h}h\cdot\dfrac1h}\right]=\tan^{-1}\dfrac{2h^3}{(h+1)^2}$$
$$\implies\lim_{h\to0^+}\dfrac{\tan^{-1}\dfrac{1+2h}h-\tan^{-1}\dfrac1h}h=\lim_{h\to0^+}O(h^2)=0$$
Finally, $$\lim_{h\to0^+}\tan^{-1}\dfrac{1+2h}h=\tan^{-1}(+\infty)=+\dfrac\pi2$$
A: I saw @mathlove's solution a different way, in case you find it hard to come up with convenient substitutions:
$(x+2)\arctan (x+2) - x \arctan(x)$ = $x(\arctan (x+2) - \arctan(x) )+2\arctan(x)$
Look at the first tirm: $x$ is going to $\infty$, but $\arctan (x+2) - \arctan(x)$ is going to $0$. And if you know the graph of $\arctan(x)$, you know the difference is going to $0$ "pretty quickly"--which is just intuition to help us know that the product is probably going to 0. 
We can prove it goes to 0 with L'Hopital:
$x(\arctan (x+2) - \arctan(x))=\frac{\arctan (x+2) - \arctan(x)}{\frac{1}{x}}$,
So
$\lim\limits_{x\rightarrow \infty} x(\arctan (x+2) - \arctan(x))=\lim\limits_{x\rightarrow\infty} \frac{\frac{1}{1+(x+2)^2}-\frac{1}{1+x^2}}{-\frac{1}{x^2}}=0$
because the $x^4$ terms cancel out of the simplified numerator.
That leaves us with $2\arctan(x)$, but it's easy to see this goes to $\pi$.
Basically the key was I saw $x$ multiplying two $\arctan$ terms, so I had the instinct to factor them. Add in some good productive staring, and voila.
