<s>[INST] Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating L'hospital] and $\underset{x \to \infty}{\lim}2. \tan^{-1}(x+2) = \pi$
so $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]= \pi$
Can anyone please  correct me If I have gone wrong anywhere?
 [/INST] Your solution is fine.
Another possible way to do it would be using that, for $t>0$,
$$\tan^{-1}(t)+\tan ^{-1}\left(\frac{1}{t}\right)=\frac \pi 2$$ making
$$A=(x+2)\tan^{-1}(x+2) -x \tan^{-1}(x)$$ $$A=(x+2)\left(\frac \pi 2-\tan ^{-1}\left(\frac{1}{x+2}\right) \right)-x\left(\frac \pi 2-\tan ^{-1}\left(\frac{1}{x}\right) \right)$$
$$A=\pi+x\tan ^{-1}\left(\frac{1}{x}\right)-(x+2)\tan ^{-1}\left(\frac{1}{x+2}\right)$$ and use, for large $t$, the equivalent
$$t\tan ^{-1}\left(\frac{1}{t}\right)\sim 1$$
If you want to go beyond, you could use Taylor expansions and get
$$A=\pi -\frac{4}{3 x^3}+\frac{4}{ x^4}+O\left(\frac{1}{x^5}\right)$$
Using your pocket calculator, for $x=10$,
$$A=12 \tan ^{-1}(12)-10 \tan ^{-1}(10)\approx 3.14058$$ while the truncated expression gives
$$\pi -\frac{7}{7500}\approx 3.14066$$
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