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Why does $$\DeclareMathOperator{arccot}{arccot}\lim_{x \to 1}\arccot\left(\frac{x^2+1}{x^2-1}\right)$$ diverge? In my textbook it says that from the positive side it's zero, and from the negative side it's $\pi$. However, when entering it into Symbolab the solution is zero (according to them, it converges). I also got $0$ in both cases, using my caveman-plug-in-the-values technique. I guess I'm solving this limit incorrectly.

What is the real solution? If I was analyzing this function and I messed up the vertical asymptote (which I'm doing in this particular problem), the rest of my analysis would be incorrect.

Can someone clear this out for me?

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  • $\begingroup$ If $x<1$ then $\frac{x^2+1}{x^2-1}<0$, so the arccot is between $\pi/2$ and $\pi$. How could the limit from the left side be $0$? $\endgroup$ – Thomas Andrews Feb 6 '16 at 21:28
  • $\begingroup$ Have you tried drawing the graph of your function on the interval (0,2)? $\endgroup$ – Bernard Masse Feb 6 '16 at 21:30
  • $\begingroup$ @ThomasAndrews It's $\pi$ from the left side. $\endgroup$ – Quant Feb 6 '16 at 21:30
  • $\begingroup$ @BernardMasse I haven't, this is a part of my analysis of the function, so I should find the asymptotes first, and everything else, before drawing it. Before doing that, I would probably make a mistake in the drawing. $\endgroup$ – Quant Feb 6 '16 at 21:31
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    $\begingroup$ We got into this issue with computer-based systems before with a different post. There are differing definitions for the range of the arc-cotangent function, so some systems (such as you used) will claim the two-sided limit exists, while a commonly-used definition (which your book uses) would say otherwise. See, for instance, mathworld.wolfram.com/InverseCotangent.html , about a quarter of the way down the page concerning the alternative conventions. $\endgroup$ – colormegone Feb 6 '16 at 21:56
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$\DeclareMathOperator{arccot}{arccot}$This is very interesting. Apparently the interpretations of $\arccot(x)$ vary! While your math textbook assumes $\arccot(x)$ as the inverse of $\cot(x)$ on $\left(0, \pi\right)$, Symbolab (and Wolfram Alpha and other mathematics software) use $\left(-{\pi \over 2}, {\pi \over 2}\right]-\{0\}$.

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The result are two different versions of $\arccot(x)$, of which the latter has a limit at $\pm \infty$ and the other has not.

So for solving such limits, you should use the definiton from your textbook. The first defintion is the common, by the way.

Edit: Wolfram has an interesting site discussing the issues with the differing definitions of $\arccot(x)$.

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  • $\begingroup$ haha very interesting! great observation $\endgroup$ – Andres Mejia Feb 6 '16 at 22:09
  • $\begingroup$ Hmm, so maybe stating the range in which we are analyzing this function would solve the problem? Thanks for an out of the box answer! $\endgroup$ – Quant Feb 6 '16 at 22:22
  • $\begingroup$ @Quant Well you should just use the definition stated in your textbook $\endgroup$ – Adrian Feb 7 '16 at 5:25
  • $\begingroup$ Nice catch. This makes me wondering why would they choose discontinuous over continuous/smooth one? $\endgroup$ – Ennar Feb 8 '16 at 17:13
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The text book is correct. Note that the limit of the argument is$\DeclareMathOperator{arccot}{arccot}$

$$\lim_{x\to 1^{\pm}}\frac{x^2+1}{x^2-1}=\pm \infty$$

and that for the Principal Values of the arccotangent, we have

$$\lim_{x\to \infty}\arccot(x)=0$$

and

$$\lim_{x\to -\infty}\arccot(x)=\pi$$

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  • $\begingroup$ This answer does not clarify why the textbook's result differ from the software's result! It just restates the textbook's solution. Worth nothing. $\endgroup$ – Adrian Feb 7 '16 at 7:37
  • $\begingroup$ @adjan It answers the exact lead question as to why the limit does not exist. It also answers the question, "What is the real solution?" It is definitely worth something. Be careful about happy trigger down voting. $\endgroup$ – Mark Viola Feb 7 '16 at 15:44
  • $\begingroup$ The real solution depends on the definition of $arccot$ which is apparently ambigious. $\endgroup$ – Adrian Feb 7 '16 at 15:47
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    $\begingroup$ The arccot function has just one range: the one from the textbook (and from wikipedia). That "other range" is just a software bug IMHO. This answer is perfect since it's categorical, and clears the confusions. $\endgroup$ – peter.petrov Feb 8 '16 at 17:12
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    $\begingroup$ @adjan OK, good to know it's intended and not a bug. That's why I said "IMHO" and not "I'm sure". So ... thanks for the info. But ... I comment whenever I want to comment. And ... in any case, the widely popular range of $arccot$ is $(0, \pi)$. $\endgroup$ – peter.petrov Feb 8 '16 at 18:34
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As far as I can tell, the different sources are giving different answers because they use different definitions of $\operatorname{arccot}(x)$. One can actually define infinitely many "branches" of $\operatorname{arccot}(x)$, one for each interval where $\cot(x)$ is one-to-one.

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Whenever $x^2-1\neq 0$, we have$\DeclareMathOperator{arccot}{arccot}$

$$ y=\arccot \left(\frac{x^2+1}{x^2-1}\right) \implies \frac{1}{\tan y}=\left(\frac{x^2+1}{x^2-1}\right) \implies y=\arctan\left(\frac{x^2-1}{x^2+1}\right) $$

So

$$ \lim_{x\to 1} y=\arctan(0)=0 $$

Note: $\arctan$ is continuous since $\tan$ is monotoinc increasing and continuous

EDIT: Okay, if you only want in terms of $\cot$, consider the graph of $\cot$ between $-\pi$ and $\pi$. $\cot(x) \to 0$ close to $x=\pm\pi$ and $\cot x \to \pm \infty$ as $x\to 0$. $\arccot$ is inverse function of $cot$. so as $x\to \pm \infty$, $\arccot x \to 0$, It doesn't depend on whether you go to $+$ or $-$ infinity, it must go to $0$.

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    $\begingroup$ Why does it actually not work to look at $\lim_{x\to1}{x^2+1 \over x^2 - 1} = \pm \infty$ and then looking how $\arccot(\pm\infty)$ behaves? $\endgroup$ – Adrian Feb 6 '16 at 21:43
  • $\begingroup$ @adjan According to wolfram alpha: wolframalpha.com/input/?i=y%3Darccot+x $\endgroup$ – user160738 Feb 6 '16 at 21:45
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$$\DeclareMathOperator{arccot}{arccot}\lim_{x \to 1}\arccot(\frac{x^2+1}{x^2-1})$$ Let set $t$ to go to 0 like this: $$t=x-1, t+1=x,x\to 1, t \to 0$$ By applying limit and using graph of $arccot(t)$ we get: $$\lim_{t\to0}\arccot(\frac{t^2+2t+2}{t^2+2t})=\arccot(\frac{2}{0})=0$$

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