Why does $\lim_{x \to 1}\operatorname{arccot}\left(\frac{x^2+1}{x^2-1}\right)$ diverge? 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? 
 A: $\DeclareMathOperator{arccot}{arccot}$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 $\left(-{\pi \over 2}, {\pi \over 2}\right]-\{0\}$.


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 $\arccot(x)$.
A: 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
$$\lim_{x\to -\infty}\arccot(x)=\pi$$
A: 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 $0$.
A: 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. 
A: $$\DeclareMathOperator{arccot}{arccot}\lim_{x \to 1}\arccot(\frac{x^2+1}{x^2-1})$$
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:
$$\lim_{t\to0}\arccot(\frac{t^2+2t+2}{t^2+2t})=\arccot(\frac{2}{0})=0$$
