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What is $\lim\limits_{z \to 0} |z \cdot \sin(\frac{1}{z})|$ for $z \in \mathbb C^*$? I need it to determine the type of the singularity at $z = 0$.

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What kind of singularity does $\sin$ have at $\infty$? –  Antonio Vargas Jan 18 '13 at 17:28
    
It is easy to compute the Laurent series of your function at $0$, so that is how to tell what kind of singularity it has there. –  GEdgar Jan 18 '13 at 17:39
    
@GEdgar That's what I did and found that the singularity at 0 must be essential. But when I tried to evaluate to this limit, I found that it had a result $< \infty$, when it should be undefined. –  StringerBell Jan 18 '13 at 19:20
    
@AntonioVargas $\lim\limits_{z \to \infty} \sin(z)$ converges, but I thought since I'm multiplying with $z$ as $z$ approaches $0$, it wouldn't matter (which - judging from the answers below - was wrong; this counts for $x \in \mathbb R$ but not in $\mathbb C$). –  StringerBell Jan 18 '13 at 19:24

2 Answers 2

up vote 5 down vote accepted

We have $a_n=\frac{1}{n} \to 0$ as $n\to \infty$ and $$ \lim_{n \to \infty}\Big|a_n\sin\Big(\frac{1}{a_n}\Big)\Big|=\lim_{n \to \infty}\frac{|\sin n|}{n}=0, $$ but $$ \lim_{n \to \infty}\Big|ia_n\sin\Big(\frac{1}{ia_n}\Big)\Big|=\lim_{n \to \infty}\frac{e^n-e^{-n}}{2n}=\infty. $$ Therefore $\lim_{z \to 0}|z\sin(z^{-1})|$ does not exist.


Notice for every $k \in \mathbb{N}$ the limit $\lim_{z\to 0}|z\sin(z^{-1})|$ does not exist. In fact $$ \lim_{n \to \infty}\Big|a_n^{k+1}\sin\Big(\frac{1}{a_n}\Big)\Big|=\lim_{n \to \infty}\frac{|\sin n|}{n^{k+1}}=0, $$ but

$$ \lim_{n \to \infty}\Big|(ia_n)^{k+1}\sin\Big(\frac{1}{ia_n}\Big)\Big|=\lim_{n \to \infty}\frac{e^n-e^{-n}}{2n^{k+1}}=\infty. $$ It follows that $z=0$ is neither a pole nor a removable singularity. Hence $z=0$ is an essential singularity.

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Can I apply the same arguments to $\lim\limits_{z \to \pi/2} \tan^2(z)$ for $z \in \mathbb C$? This limit would also be undefined. –  StringerBell Jan 18 '13 at 21:11
    
Yes, and you'll find that the limit $\lim_{z \to \pi/2}(z-\pi/2)^k\tan^2x$ exists for $k=4$, i.e. $z=\pi/2$ is a pole of order $4$ for $\tan^2$. –  Mercy Jan 18 '13 at 21:34
    
This, I don't understand - but I have opened a new question for that (in case your interested: math.stackexchange.com/questions/281647/…) thanks already! –  StringerBell Jan 18 '13 at 21:58
    
sorry, the order is $2$, not $4$! –  Mercy Jan 18 '13 at 22:33

It is undefined. Consider the function $g(z)=\frac{\sin{z}}{z}$. This can be made an entire function by setting $g(0)=0$. But then you are looking for $\lim_{z\to\infty} g(z)$. Any non-polynomial entire function must have an essential singularity at $\infty$, so this limit does not exist.

Alternatively:

For $x$ real, $\lim_{x\to\infty} g(x)$ is clearly zero, since $|\sin x|\leq 1$.

But when $z=ix$ is imaginary, then $g(z)=\frac{e^x-e^{-x}}{2x}$, which has infinite limit as $x\to\infty$.

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Why must any non-constant entire function have an essential singularity at $\infty$? Because the "value" $\infty$ is technically not "allowed" for functions? –  StringerBell Jan 18 '13 at 19:15
    
Hmmm, $f(z)=z$ is a non-constant entire function and it clearly does not have an essential singularity at infinity, does it? –  Nils Matthes Jan 18 '13 at 19:26
    
@NilsMatthes Sorry, you are correct, it only has an essential singularity of the function is not a polynomial. –  Thomas Andrews Jan 18 '13 at 19:31
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@StringerBell I was slightly wrong, it is only true for non-polynomial entire functions that the singularity is essential. But check the Wikipedia page for Entire function: en.wikipedia.org/wiki/Entire_function –  Thomas Andrews Jan 18 '13 at 19:33
    
@ThomasAndrews: I think you mean more generally rational function, not polynomial. –  Nils Matthes Jan 18 '13 at 19:33

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