Evaluate $\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)$ 
$\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)$

What are the good/ clever methods to evaluate this limit? 
I tried taking $\tan^{-1} (x+5) = \theta$ to avoid inverse functions but its not helpful and makes it even more complicated. 
I also tried $\tan^{-1}a - \tan^{-1}b$ formula for the terms attached to x but that does not help to get rid of other terms multiplied by $1$ and $5$.
Edit: (Please address this in your answer) 
Can't we directly do this: 
$\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)$
$= (x+5)\dfrac{\pi}{2} - (x+1)\dfrac{\pi}{2}$
$ = \dfrac {5\pi - \pi}{2} = 2\pi$
I don't see anything wrong with it and it gives the right answer. 
Is this method correct? Can it be used in other questions too? 

Use Taylor expansion:
$$\tan^{-1}(x+1)=\frac{\pi}{2} - \frac 1x + \frac 1{x^2} - \frac{2}{3 x^3} + O\left(\frac{1}{x^5}\right);\\
\tan^{-1}(x+5)=\frac{\pi}2 - \frac 1x + \frac5{x^2} - \frac{74}{3 x^3} + \frac{120}{x^4} + O\left(\frac 1{x^5}\right);\\
\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)=\\
\lim_{x\to \infty} (x+5)\left[\frac{\pi}{2}-\frac1x+\frac{5}{x^2}+O\left(\frac 1{x^3}\right)\right]- (x+1)\left[\frac{\pi}{2}-\frac1x+\frac{1}{x^2}+O\left(\frac 1{x^3}\right)\right]=\\
\lim_{x\to \infty} \left[2\pi-\frac 4x+O\left(\frac1{x^2}\right)\right]=2\pi.$$
