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? 
 A: The expression under limit can be written as $$x\{\tan^{-1}(x+5)-\tan^{-1}(x+1)\}+\{5\tan^{-1}(x+5)-\tan^{-1}(x+1) \}$$ As you have noted in your question the second term tends to $5\pi/2-\pi/2=2\pi$. The first term on the other hand can be written as $$x\tan^{-1}\frac{4}{1+(x+1)(x+5)}=x\tan^{-1}t\text{ (say)} $$ where $t\to 0$. Noting that $(1/t)\tan^{-1}t\to 1$ we have $$x\tan^{-1}t=xt\cdot\frac {\tan^{-1}t}{t}\to 0\cdot 1=0$$ as $xt=4x/(1+(x+1)(x+5))\to 0$. Thus the desired limit is equal to $2\pi$.

Your approach has a serious problem as you can't replace a part of the expression with its limit in general. See this answer for more details.
You can also look at it in this way. Suppose the question is modified to evaluate the limit of $$(x^2+5)\tan^{-1}(x+5)-(x^2+1)\tan^{-1}(x+1)$$ If we proceed as per your approach we again get the answer as $2\pi$. But the right answer here would be $2\pi+4$. Proceeding as I have explained above you will get $2\pi$ plus a term $x^2\tan^{-1}t$ and this will lead to $x^2t$ which tends to $4$.
A: Hint: $(x+5)\tan^{-1}(x+5) - (x+1)\tan^{-1}(x+1) = \dfrac{\tan^{-1}(x+5)-\tan^{-1}(x+1)}{\frac{1}{x}}+ 5\tan^{-1}(x+5) - \tan^{-1}(x+1)$. Use L'hopitale rule on the first term and the other terms have well-known limits....
A: Use Taylor expansion:
$$\tan^{-1}(x+1)=\frac{\pi}{2} - \frac 1x + \frac 1{x^2} - \frac{2}{3 x^3} + O\left(\frac{1}{x^5}\right);\\
\tan^{-1}(x+5)=\frac{\pi}2 - \frac 1x + \frac5{x^2} - \frac{74}{3 x^3} + \frac{120}{x^4} + O\left(\frac 1{x^5}\right);\\
\lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)=\\
\lim_{x\to \infty} (x+5)\left[\frac{\pi}{2}-\frac1x+\frac{5}{x^2}+O\left(\frac 1{x^3}\right)\right]- (x+1)\left[\frac{\pi}{2}-\frac1x+\frac{1}{x^2}+O\left(\frac 1{x^3}\right)\right]=\\
\lim_{x\to \infty} \left[2\pi-\frac 4x+O\left(\frac1{x^2}\right)\right]=2\pi.$$
A: Consider
$$
f_a(t)=\arctan\dfrac{1+at}{t}
$$
defined for $t>0$; then $\lim_{t\to0^+}f_a(t)=\pi/2$. Also, for $t>0$,
$$
f_a'(t)=-\frac{1}{t^2}\frac{1}{1+\dfrac{(1+at)^2}{t^2}}=-\frac{1}{t^2+(1+at)^2}
$$
and therefore $\lim_{t\to0^+}f_a'(t)=-1$.
With the substitution $x=1/t$, your limit can be rewritten as
$$
\lim_{t\to0^+}\biggl(5f_5(t)-f_1(t)+\frac{f_5(t)-f_1(t)}{t}\biggr)
$$
With l'Hôpital and the computation above, the limit of the fraction is $0$, so the limit is
$$
\frac{5\pi}{2}-\frac{\pi}{2}=2\pi
$$
Your method is incorrect: you cannot write
$$
\lim_{x\to\infty}\bigl((x+5)\arctan(x+5)- (x+1)\arctan(x+1)\bigr)
= (x+5)\dfrac{\pi}{2} - (x+1)\dfrac{\pi}{2}
$$
because the limit cannot depend on $x$ and you're essentially using $\infty-\infty=0$, which is incorrect.
