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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'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 )2012-09-14

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

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    Thak boss ami r parchiiina2013-05-08