Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to find the limit $$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \left( \sin^n \left( \frac{x}{2} \right) + \left( \frac{1}{\pi} \cdot \operatorname{arccot}(x) \right)^n \right)$$ Where $\operatorname{arccot}(x)$ is defined like this:


And the answer is obviously something like this $$\lim_{n \to \infty} f_n(x) = \begin{cases} 1, \quad \text{when } x \in \{ \, (1+4k) \pi \mid k \in \mathbb{Z} \, \}; \\ \text{non-existent}, \quad \text{when } x \in \{ \, (-1+4k) \pi \mid k \in \mathbb{Z} \, \}; \\ 0, \quad \text{otherwise}; \end{cases}$$

But what is the meaning of $\operatorname{arccot}(x)$ in this situation? My limit doesn't really assess it in any way...

share|cite|improve this question
up vote 0 down vote accepted

Hint: the range of $\operatorname{arccot}$ is $(0, \pi)$, if I'm not mistken, so $|\operatorname{arccot}(x)/\pi| < 1$ for any real $x$. That takes care of the second piece. For the first piece, there are three cases, because $\sin(x/2)$ can be $-1$, $1$, or strictly between $-1$ and $1$.

The function $\operatorname{arccot}$ is a lot less commmonly used than $\tan^{-1}$. I am pretty sure $\operatorname{arccot}(x)$ is the (unique) angle $\theta$ in $(0, \pi)$ with $\cot \theta = x$.

share|cite|improve this answer

The link you gave had the range of $arccot(x)$: $$0 \lt arccot(x) \lt \pi$$ Then: $$0 \lt \frac{1}{\pi}arccot(x) \lt 1$$ Now consider again your problem and you'll see that: $$ \lim_{n\to\infty}{(\frac{1}{\pi}arccot(x))^n} = 0 $$

share|cite|improve this answer
But apparently $\lim_{n \to \infty} \lim_{x \to -\infty} \left( \frac{1}{\pi} \operatorname{arccot} (x) \right)^n = 1$ and that confuses me. – Pranasas Apr 12 '13 at 22:54
@Pranasas: when you define $f(x)$, you let $n \to \infty$ while x stays the same – Stefan Smith Apr 13 '13 at 0:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.