Q:

$ \lim_{x\rightarrow 0^-} \frac{\operatorname{arccot}(x) - \frac{\pi}{2}}{x}$

$$\lim_{x\rightarrow 0^-} \frac{\operatorname{arccot}(x) - \frac{\pi}{2}}{x}$$
The title says everything. I already know the limit is $+\infty$, I just want to see how it can be calculated. (Please don't use L'Hôpital's rule, I haven't covered it yet at school)

A:

At first a plot to show the limit is not -1.

The limit of $\operatorname{arccot}(x)$ as $x\to 0$ and $x<0$ is $-\frac{\pi}{2}$.
To see that, we use that $\operatorname{arccot(x)}=\frac{1}{\arctan\left(\frac{1}{x}\right)}$
and hence
$$\underset{x<0}{\lim_{x\to 0}} \operatorname{arccot}(x)=\lim_{x\to -\infty} \arctan(x)=-\frac{\pi}{2}$$
So your limit is the same as 
$$-\underset{x<0}{\lim_{x\to 0}}\frac{\pi}{x}$$
This is the same as 
$$\pi \cdot \underset{x>0}{\lim_{x\to 0}} \frac{1}{x}$$

