<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] Here is another straightforward way to obtain the limit.
Using the integral relation 
$$\arctan u=\int_0^u{dt\over1+t^2}$$ 
we have
$$\begin{align}
(x+2)\arctan(x+2)-x\arctan x
&=2\arctan(x+2)+x\int_0^{x+2}{dt\over1+t^2}-x\int_0^x{dt\over1+t^2}\\
&=2\arctan(x+2)+x\int_x^{x+2}{dt\over1+t^2}
\end{align}$$
Now $2\arctan(x+2)\to\pi$ as $x\to\infty$, while
$$0\le x\int_x^{x+2}{dt\over1+t^2}\le x\int_x^{x+2}{dt\over x^2}={2\over x}\to0$$
Thus
$$\lim_{x\to\infty}((x+2)\arctan(x+2)-x\arctan x)=\pi$$
</s>