<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 but it can be shortened. By Lagrange's theorem $(x+2)\arctan(x+2)-x\arctan(x)$ is twice the value of the derivative of $z\arctan z$ at some $z\in(x,x+2)$. Since $\frac{d}{dz}z\arctan(z) = \underbrace{\frac{z}{1+z^2}}_{\to 0}+\underbrace{\arctan(z)}_{\to \pi/2}$, the wanted limit is $\pi$.
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