# Limit on a spiral

I was thinking about limits of functions along various spirals and this one stumped me a bit. The limit that needs to be found is ultimately:

$$\lim_{\varphi\to\infty} \coth\varphi\csc\varphi$$

Here is the sprial that it comes from:

The equation for this is $(x,y)=(\tanh \varphi \sin \varphi, \tanh \varphi \cos \varphi)$ and the function for which I desire the limit is $f(x,y)=\frac{1}{x}$ I'm interested in the limit as $\varphi\to\infty$.

I've got an answer using computer algebra, which is

$$L=\lim_{\varphi\to\infty} \coth\varphi\csc\varphi \notin (-1,1)$$

I have no intuitive understanding of why it isn't just $L\in[-\infty,\infty]$.

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I think the limit does not exist: try a plot. – Tony Piccolo Jun 15 '13 at 7:13
Maple:$\quad$plot(coth(phi)*csc(phi),phi=-20..20,-20..20,numpoints=100,discont=t‌​rue); . – Tony Piccolo Jun 15 '13 at 11:11

## 3 Answers

It is clear that the range of $$\coth(\varphi)\csc(\varphi)=\frac {e^{2\varphi}+1}{(e^{2\varphi}-1)\sin(\varphi)}$$ is $(-\infty,-1) \cup (1,\infty).$ Therefore, no partial limit of the function under consideration belongs to $[-1,1]$. On the other hand, for every $a \in (-\infty,-1) \cup (1,\infty)$ and for every $E>0$ the intermediate value theorem implies that the equation $$\sin\varphi= \frac {e^{2\varphi}+1}{(e^{2\varphi}-1)a}$$ has a solution $\varphi_0 > E.$ The last statement means that $a$ is a partial limit of the function under consideration.

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What do you mean partial limit, do you have a link? – Lucas Jun 15 '13 at 8:20
It seems to me that you have shown that, if $a$ belongs to the range of $z=\coth(\varphi) \csc(\varphi) \;$ (by the way are you sure about $1$ and $-1$ ?), then in every neighborhood of $\infty$ there exists a point $\varphi_0$ such that $z(\varphi_0)=a$, but I don't know what do you mean by partial limit. – Tony Piccolo Jun 16 '13 at 10:20

Hint: The range of $\coth \varphi$ is $(-\infty,-1)\cup(1,\infty)$. The range of $\csc \varphi$ is $(-\infty,-1]\cup[1,\infty)$.

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This leads me to think $\{1,-1\}$ should be excluded. Though $\lim_{x\to\pm\infty} \coth x = \pm1$, is probably why it should be included. – Lucas Jun 15 '13 at 7:26
@Lucas Try $\lim_{\phi \to \infty} \csc (\phi)$ . – Tony Piccolo Jun 15 '13 at 7:42
@Lucas: As Tony has indicated the limit does not exist. Morally the function is $\csc \varphi$ in the limit for the reason you mention. – user26872 Jun 15 '13 at 18:47
@oen Morally? Is that an autocomplete error? – Lucas Jun 15 '13 at 19:04
@Lucas: It is a figure of speech sometimes used in mathematics and physics. – user26872 Jun 15 '13 at 19:26

Let $$z=\coth\varphi\csc\varphi$$I explain why $$\lim_{\varphi \to \infty}z$$does not exist.
As one can prove easily, z is undefined for $\varphi=k\pi \,$ with $\, k \in \Bbb Z$.
The function has vertical asymptotes in these points.
The one-sided limits in these points are infinite but different (except $0$), depending on the direction (left or right). For example $$\lim_{\varphi \to \pi -} z=\infty$$ but $$\lim_{\varphi \to \pi +} z=- \infty$$In an informal language z assumes in every neighborhood of $\infty$ great and small at will values.
A standard calculus exercise: that's all.
I suggest, in these cases, a graphical approach at first (see my Maple code).

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