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?

Yes, the solution is correct. 
You are using:
$$\arctan a\pm \arctan b=\arctan\frac{a\pm b}{1\mp ab}$$
to simplify:
$$x\tan^{-1}(x+2)-x\tan^{-1}x=x\tan^{-1}\frac{(x+2)-x}{1+(x+2)x}=x\tan^{-1}\frac{2}{1+2x+x^2}.$$
However, to evaluate the limit you can avoid using the L'Hospital's rule:
$$\lim_{x \to \infty} x \tan^{-1} \frac {2}{1+2x+x^2}=
\lim_{x\to\infty} \frac{\tan^{-1}\frac{2}{1+2x+x^2}}{\frac{2}{1+2x+x^2}}\cdot \frac{\frac{2}{1+2x+x^2}}{\frac{1}{x}}=1\cdot 0=0,$$
where $\lim_\limits{x\to 0} \frac{\tan^{-1}x}{x}=1$.
