A Limit Problem

**bkarpuz** · February 9th 2010, 07:39 AM

Dear friends,

I need help in showing the following.
$\lim_{\substack{\lambda\in\mathbb{R}\\ \lambda\to0}}\bigg(\frac{1}{\lambda}\log|1+z\lambd a|\bigg)=\mathrm{Re}(z)$

Thanks!

bkarpuz

**Laurent** · February 9th 2010, 07:57 AM

Originally Posted by bkarpuz

Dear friends,

I need help in showing the following.
$\lim_{\lambda\to0}\bigg(\frac{1}{\lambda}\log|1+z\ lambda|\bigg)=\mathrm{Re}(z)$

Thanks!

bkarpuz

(you should specify that $\lambda\in\mathbb{R}$ )

After expanding you get $|1+z\lambda|^2=1+2{\rm Re}(z)\lambda+o(\lambda)$ when $\lambda\to 0$ , $\lambda\in\mathbb{R}$ , from which the result follows quickly using $\log (1+u)=u+o(u)$ when $u\to 0$ , and $\log u=\frac{1}{2}\log u^2$ .

**bkarpuz** · February 9th 2010, 12:02 PM

Originally Posted by Laurent

(you should specify that $\lambda\in\mathbb{R}$ )

After expanding you get $|1+z\lambda|^2=1+2{\rm Re}(z)\lambda+o(\lambda)$ when $\lambda\to 0$ , $\lambda\in\mathbb{R}$ , from which the result follows quickly using $\log (1+u)=u+o(u)$ when $u\to 0$ , and $\log u=\frac{1}{2}\log u^2$ .

How did I miss this?! :S
Thanks for your reply Laurent, it has been a long time not heard from you. :]

Actually I am working with the quotient when $\lambda\in\mathbb{C}$ and its making me confused! :S

**bkarpuz** · February 9th 2010, 09:44 PM

Originally Posted by bkarpuz

How did I miss this?! :S
Thanks for your reply Laurent, it has been a long time not heard from you. :]

Actually I am working with the quotient when $\lambda\in\mathbb{C}$ and its making me confused! :S

Actually, the limit can be computed when $\mathrm{Re}(\lambda)\neq0$ .
In this case, we have $\lambda=r\mathrm{e}^{i\theta}$ with $\theta\neq\pm\pi/2$ .
So that, for the function $f(z):=|z|^{-1}\log|1+z|$ for $z\in\mathbb{C}\backslash\{-1\}$ , we have
$\lim_{r\to0^{+}}f(r\mathrm{e}^{i\theta})=\lim_{r\t o0^{+}}\frac{1}{2r}\log\big(1+2r\cos(\theta)+r^{2} \big)=\cos(\theta)$
by using the fact mentioned previously by Laurent ( $\lim\nolimits_{\lambda\in\mathbb{R},\ \lambda\to0}f(\lambda)=1$ ).
On the other hand if $\mathrm{Re}(\lambda)=0$ , it can be easily computed as
$\lim_{r\to0^{+}}f(r\mathrm{e}^{\pm i\pi/2})=\lim_{r\to0^{+}}\frac{1}{2r}\log\big(1+r^{2}\b ig)=0.$

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