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

Question

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

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)

Please, avoid double dollar signs in titles and include your question in the body of the post, not only in the title. — Pedro, Mar 03 '13 at 19:11
The expression is by definition the (leftsided) derivative of $\arccot(x)$ at $x=0$. By the ways, this makes the result $-1$, not $+\infty$. — Hagen von Eitzen, Mar 03 '13 at 19:14
@HagenvonEitzen The limit of arccot$(x)$ when $x\to0$, $x\lt0$, is $-\pi/2$, not $+\pi/2$, hence there is no derivative here and the result is not $-1$. — Did, Mar 03 '13 at 19:17
@Did: What can we do when we find the plot of the function admits Hagens point? Thanks — Mikasa, Mar 03 '13 at 19:33
@HagenvonEitzen Let me qualify my last statement: there is apparently some variations in the definition of arccot. My comment is based on W|A definition and yours on WP's definition. The rest is theology. — Did, Mar 04 '13 at 07:24

Dominic Michaelis · Accepted Answer · 2013-03-04T13:15:20.977

1

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

enter image description here

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 $$\pi \cdot \underset{x>0}{\lim_{x\to 0}} \frac{1}{x}$$

edited Mar 04 '13 at 13:15

answered Mar 03 '13 at 19:21

Dominic Michaelis

19,935

Thanks, very good explanation, I actually understood it. – Tedy S. Mar 04 '13 at 20:00

Emanuele Paolini · Answer 2 · 2013-03-04T06:55:13.457

0

I assume that $\mathrm arccot x = \pi/2 - \arctan x$. Try to recover the definition of derivative for the arccot (or better: the $\arctan$) function:

$$ \lim_{x\to 0^-}\frac{\mathrm {arccot}x - \pi/2}{x} = \lim_{x\to 0^-} \frac{-\arctan x}{x} = - \arctan'(0) = -1. $$

edited Mar 04 '13 at 06:55

answered Mar 03 '13 at 19:14

Emanuele Paolini

21,447

the derivatives of arccot and arctan are bounded and hence won't get near infty anywhere – Dominic Michaelis Mar 03 '13 at 19:43
We don't agree on the definition of arccot (quite a not useful function). I took the definition given in wikipedia: http://en.wikipedia.org/wiki/Inverse_trigonometric_functions – Emanuele Paolini Mar 04 '13 at 06:58
There seems to be some variations in the definition of arccot. – Did Mar 04 '13 at 07:20

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

2 Answers2