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

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....


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


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 $4$.

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