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

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?

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    $\begingroup$ Maybe $f-g=\frac{1/g-1/f}{1/fg}$ and use L'Hôpital Rule? $\endgroup$ – Tito Eliatron Dec 23 '18 at 17:19
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    $\begingroup$ is it arctan((x+5)-(x+1))arctan...... or is it arctan(x-5)((x+1)arctan(x+1)) $\endgroup$ – ricky Dec 23 '18 at 17:29
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    $\begingroup$ No, the method is incorrect: the limit cannot depend on $x$. $\endgroup$ – egreg Dec 23 '18 at 17:43
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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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    $\begingroup$ But other Answers say second term cannot be taken as 2pi just like that $\endgroup$ – Archer Dec 23 '18 at 19:12
  • $\begingroup$ @Abcd: why not? You can use limit laws to do so. $\endgroup$ – Paramanand Singh Dec 23 '18 at 19:14
  • $\begingroup$ @Abcd: see last paragraph of my updated answer. $\endgroup$ – Paramanand Singh Dec 23 '18 at 19:17
  • $\begingroup$ @Abcd: you should carefully note the difference between your approach which is not permitted by limit laws and my handling of second term which looks similar but is permitted by the same laws. $\endgroup$ – Paramanand Singh Dec 23 '18 at 19:21
  • $\begingroup$ Which laws??... $\endgroup$ – Archer Dec 23 '18 at 19:31
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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....

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  • $\begingroup$ Please see the edit. $\endgroup$ – Archer Dec 23 '18 at 17:34
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Use Taylor expansion: $$\tan^{-1}(x+1)=\frac{\pi}{2} - \frac 1x + \frac 1{x^2} - \frac{2}{3 x^3} + O\left(\frac{1}{x^5}\right);\\ \tan^{-1}(x+5)=\frac{\pi}2 - \frac 1x + \frac5{x^2} - \frac{74}{3 x^3} + \frac{120}{x^4} + O\left(\frac 1{x^5}\right);\\ \lim_{x\to \infty} (x+5)\tan^{-1}(x+5)- (x+1)\tan^{-1}(x+1)=\\ \lim_{x\to \infty} (x+5)\left[\frac{\pi}{2}-\frac1x+\frac{5}{x^2}+O\left(\frac 1{x^3}\right)\right]- (x+1)\left[\frac{\pi}{2}-\frac1x+\frac{1}{x^2}+O\left(\frac 1{x^3}\right)\right]=\\ \lim_{x\to \infty} \left[2\pi-\frac 4x+O\left(\frac1{x^2}\right)\right]=2\pi.$$

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    $\begingroup$ Please see the edit. $\endgroup$ – Archer Dec 23 '18 at 17:34
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    $\begingroup$ Formally, you can not evaluate it partially, because it is the indeterminate form $\infty-\infty$. As a counterexample, consider $\tan^{-1}(x^2+1)$ in the second term, then the limit will be $2\pi-1$. $\endgroup$ – farruhota Dec 23 '18 at 17:41
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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.

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