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Let $f(z)= \dfrac 1 {e^{1/z}+1}$. Does this have a removable singularity at $z=\infty$?

I found that all the singulairties (other than $\infty$) are bounded. So $z=\infty$ is a isolated singularity of $f(z)$. But does $f(1/z)=\dfrac 1 {e^{1/(1/z)}+1}=\dfrac 1 {e^{z}+1}$ have removable singularity at $z=0$ because just it has $z$ in the denominator?

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I don't see any singularity at all. Certainly $1/(e^z+1)$ has no singularity at $z=0$. –  Gerry Myerson Sep 14 '12 at 2:58
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@Gerry: That's true in the sense that a removable singularity is not a singularity at all, but e.g., $z\mapsto z$ has a removable singularity at $0$ if its domain is specified to be $\mathbb C\setminus \{0\}$, and I believe $f$ is implicitly assumed in this problem takes inputs from a subset of $\mathbb C$, hence the domain excludes $\infty$. –  Jonas Meyer Sep 14 '12 at 3:02
    
I'm not sure but perhaps $\,\infty\,$ is automatically considered "a singularity" in every case by some authors ( I know we did so while studying this stuff ) –  DonAntonio Sep 14 '12 at 3:04

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up vote 4 down vote accepted

Yes, it is removable: in fact $\lim_{z \to \infty} f(z) = 1/2$.

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$f(z)$ has a removable singularity at $z=a$ iff $\lim_{z\to a} (z-a)f(z)=0$,So how did you conclude that? –  El Angel Exterminador May 7 '13 at 14:45
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As $z \to \infty$, $1/z \to 0$, $e^{1/z} \to e^0 = 1$, etc. –  Robert Israel May 8 '13 at 6:14
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@Tsotsi: Look at Riemann's Theorem‌​. Your characterization of a removable singularity only works for finite $a$. In his last comment, Robert Israel is using characterization 2: $f$ is continuously extendable over $a$. –  robjohn May 8 '13 at 9:24
    
Thak boss ami r parchiiina –  El Angel Exterminador May 8 '13 at 9:44

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