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I've read the following claim, and wander how to prove or disprove it:

For a given function $f(z)=g(z)\cdot h(z)$ so that $\lim_{z\to0}g(z)=a$,

there is $h(z)$ so that $f(z)=g(z)\cdot [a+h(z)]$

and $\lim_{z\to0}h(z)=0$

Can someone explain that?

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I think your question is confused. Please be careful with which function is $g$ and which is $h$. I suspect what you mean is, if $\lim_{z \to 0} g(z) = a$, then you can write $g(z) = a + r(z)$ with $\lim_{z \to 0} r(z) = 0$, and $g(z) h(z) = (a + r(z)) h(z)$. – Robert Israel Feb 17 '12 at 17:50
My question is wrong. – Michael Feb 17 '12 at 18:07
up vote 1 down vote accepted

If I understand you correctly then this statement is false: $$\lim_{z\to 0}g(z)=a$$ $$\lim_{z\to 0}h(z)=0$$ Which means $$\lim_{z\to 0}f(z)=\lim_{z\to 0}g(z)\lim_{z\to 0}h(z)=0\neq \lim_{z\to 0}f(z)=\lim_{z\to 0}g(z)\lim_{z\to 0}[a+h(z)]=a^2$$

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Robert comment is true. My question was confused. Sorry for that. – Michael Feb 17 '12 at 18:17
One more uncertainty: is the question about proving/disproving $f(z)=g(z)h(z)=(a+r(z))h(z)$ or $\lim_{z\to 0}f(z)=\lim_{z\to 0}g(z)h(z)=\lim_{z\to 0}(a+r(z))h(z)$ ? – Julius Feb 17 '12 at 18:41

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