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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:

arccot(x)

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...

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2 Answers 2

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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 $\cot \theta = x$.

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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: $$ \lim_{n\to\infty}{(\frac{1}{\pi}arccot(x))^n} = 0 $$

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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
1  
@Pranasas: when you define $f(x)$, you let $n \to \infty$ while x stays the same –  Stefan Smith Apr 13 '13 at 0:27

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