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