I was calculating a $$\lim_{{x \to -\infty}}\left(\frac{4-x}{1-x}\right)^{2x\mathrm{arccot}(x)}$$ and found a problem og how wolframalpha presents $\mathrm{arccot}(x)$. I was taughted that plot of arccot(x) looks like this:
But wolframalpha shows it like this
That why we have completely different answers. Because according to my representation $$\lim_{{x \to -\infty}}\mathrm{arccot}(x)$$ equals to Pi, when wolfram says it equals to 0.
I calculated that the answer is $$e^{6\pi}$$, answer of wolfram is $1$. Please help me understand where is a mistake.
Recall that if $y = \cot^{-1} x$, this means that $y$ is an angle whose cotangent is $x$; i.e., $$\cot y = x.$$ Due to the periodicity of the circular trigonometric functions, it is clear that for a given value of $x$, there are infinitely many values of $y$ satisfying the above relationship; namely, $$\cot(y + k\pi) = x$$ for any integer $k \in \mathbb Z$. Consequently, it is misleading to speak of "the" inverse cotangent of a number, although we tend to do this rather frequently. In a similar vein, we also use the phrase "the square root" but in truth, there are in general two square roots of a positive real number. While there are reasons to assume that by "the square root" we mean the positive square root, there is not necessarily a similar canonical choice for the inverse cotangent.
One reason to choose the codomain of $f(x) = \cot^{-1} x$ as $f : \mathbb R \to (0,\pi)$ is that it ensures the continuity of $f$ over the real line, as you can see from the graph you plotted (blue line). However, it is not the only possible choice: Mathematica chooses the mapping so that the codomain of the inverse cotangent is the same interval as that for the inverse tangent: $f(x) \in (-\pi/2, \pi/2]$. This choice has to do with notions of branch cuts in complex analysis but a discussion of this is not within the scope of this answer.
One way to address this discrepancy is to avoid making the choice at all: that is to say, if we regard the mapping $f(x) = \cot^{-1} x$ as that of a number to a set, what would be the outcome of the limit as a set of values? The values $e^{6\pi}$ and $1$ that you and Mathematica obtained can be regarded as elements of that set. Another way is to simply qualify your choice of codomain when evaluating the limit--that is to say, explain which element of the set of all possible values is being chosen.