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.

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, $$\lim_{h\to0^+}\tan^{-1}\dfrac{1+2h}h=\tan^{-1}(+\infty)=+\dfrac\pi2$$
