# Thread:

1. ## A Limit Problem

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

2. 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 $\log u=\frac{1}{2}\log u^2$.

3. 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

4. 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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