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?

Using the Mean Value Theorem, the following statement holds:
For $f(z) = 2z\tan^{-1}{z}$
$$f'(x+\xi)=\frac{2(x+2)\tan^{-1}{(x+2)}-2x\tan^{-1}{x}}{x+2-x}$$
For some $\xi \in (0,2)$
The left hand side simplifies to:
$$f'(x+\xi)=2\tan^{-1}{(x+\xi)}+\frac{2(x+\xi)}{1+(x+\xi)^2}$$
The right hand side simplifies to the limit you seek.
By taking the limit of the left hand side, the limit is clearly $\pi$
