I need advice on calculating this limit of a function resulting in $-\frac14\pi$

Question

I am looking for advice on solving this limit of a function.

I am struggling to find the correct process:

$$\lim\limits_{x\to-\infty}\operatorname{arccotg}\frac{x}{(x^2-4)^{\frac12}}$$

Answer 1

Note that $\text{arccotg}$ is the inverse function of $\cot$. (I'm using whichever version always outputs within the range $[-\pi/2, \pi/2]$.)

The $\operatorname{arccotg}$ function is continuous near infinity (and therefore respects limits), so the answer is:

$$\operatorname{arccotg} \lim_{x \to -\infty} \frac{x}{\sqrt{x^2-4}}$$

That limit expression is (by L'Hôpital)

$$\lim_{x \to -\infty} \frac{1}{1/2 (x^2-4)^{-1/2} 2x} = \lim_{x \to -\infty} \frac{\sqrt{x^2-4}}{x} = \lim_{x \to \infty} -\sqrt{1-\frac{4}{x^2}}$$

By continuity of the square root near infinity, that is

$$-\sqrt{\lim_{x \to \infty} \left( 1-\frac{4}{x^2} \right)} = -1$$

So the entire answer is $\operatorname{arccotg}(-1)$, which by oddness is $-\text{arccotg}(1) = -\frac{\pi}{4}$.

Answer 2

With equivalents, one sees at once that $$\frac x{(x^2-4)^{\tfrac12}}\sim_{-\infty}\frac x{\lvert x\rvert}\to -1,$$ hence by continuity, $$\operatorname{arccot}\frac x{(x^2-4)^{\tfrac12}}\enspace\text{tends to}\enspace\operatorname{arccot}(-1)=\frac{3\pi}4. $$ as the range of $\operatorname{arccot}$ is $(0,\pi)$.