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Any idea about this problem:

$$\lim_{z\to0}\frac{\log(1+z)}{z}=1$$

Any hints would be appreciated.

felipeuni
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  • Logs with complex numbers are tricky: which branch did you choose for it? – Avitus Sep 04 '13 at 08:06
  • @Avitus $\mathbb{C}^-=\mathbb{C}\setminus {z\in \mathbb{R}:z\le 0}$ – felipeuni Sep 04 '13 at 08:15
  • @StefanHansen Even though the question is pretty much the same, this on is tagged with complex analysis and the other is tagged with calculus and limits. Even though this works here it is not immeadetely clear, why the existence of the limit in the restriction on $\mathbb{R}$ gives the existence of the limit in $\mathbb{C}$. In many cases this is just wrong – Dominic Michaelis Sep 04 '13 at 08:41
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    @DominicMichaelis: Point taken, I did not take notice of the tag. I was merely focusing on the $\lim_{z\to 0}$ which suggested that $z$ is a real parameter. Sorry for the noise. – Stefan Hansen Sep 04 '13 at 08:54

5 Answers5

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Hint:

$$\frac{\log(1+z)}{z}=\frac{\log(1+z)-\log(1)}{(z+1)-1}$$ So the limit is equal to $\log'(1)=1$

Dominic Michaelis
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I think if it is in the reals,

Use z = 1/n , and consider what happens when n----> infinity.

:Log(1+z)/z would become log (1+ [1/n])^n .

It is easy to see that the latter expression tends to log e =1, because as n ----> infinity , (1+[1/n])^n -----> e.

Yuanny
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Hint: You can use that the Laurent expansion of the numerator is $$ \log(1+z)=\sum_{n=1}^\infty (-1)^{n+1}\dfrac{z^n}{n} \qquad\qquad \text{for $|z|<1$} $$

Andrea Orta
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Use l'Hopital rule,

$$ \lim_{z\to0}\frac{\log(1+z)}{z}=\lim_{z\to0}\frac{(\log(1+z))'}{(z)'}=\lim_{z\to0}\frac{1}{1+z}=1 $$ For complex functions, L'Hopital rule does not work in general but works in special cases. Because your functions are analytic at 0, the rule works here.

Arash
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If I right, there is a l'Hôpital's rule for a complex function, and hence $$\lim_{z\to 0}\frac{\log(1+z)}{z}=\lim_{z\to 0}\frac{\frac{1}{1+z}}{1}=1.$$

Mher
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