$\underset{z\rightarrow \infty }{\lim} \, \frac{z}{\log z}$

Question

I want to calculate $$\underset{z\rightarrow \infty }{\lim} \, \frac{z}{\log z}=\underset{z\rightarrow \infty }{\lim} \,\frac{z}{\log|z|+ i\operatorname{Arg} z} $$

but I have some problems

First $\underset{z\rightarrow \infty }{\lim} \,\operatorname{Arg}z$ doesn't exist, that means that the denominator is $\infty$ because of the $\log |z|$?

If we don't have a problem with the denominator and its just $\infty $ how can I solve the $\frac{\infty}{\infty}$ problem

I tried to take the norm $| \,\frac{z}{\log|z|+ i\operatorname{Arg} z} |<\frac{|z|}{|\log|z|-i\pi|}$ or to show that the limit doesn't exist by taking 2 subsequences ect, but it didn't help me solve the $\frac{\infty}{\infty}$ problem.

Any hints?

Answer 1

$\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}}\gae{Log}\gae{Arg}$If you are taking the limit in the Riemann sphere $\Bbb C\cup\{\infty\}$ and $\Log$ is some banch cut of the logarithm (let's say, $\Arg$ takes values in $[0,2\pi)$), then for all $\lvert z\rvert>e^{2\pi}$ you have $$\frac{\lvert z\rvert}{\ln\lvert z\rvert+2\pi}\le\left\lvert \frac{z}{\Log z}\right\rvert\le \frac{\lvert z\rvert}{\ln\lvert z\rvert-2\pi}$$

which diverges to $(+)\infty$ as $\lvert z\rvert\to \infty$ because it's a mundane real limit in the real variable $\lvert z\rvert$.

Therefore, under the assumptions I have mentioned, $\lim_{z\to\infty}\frac z{\Log z}=\infty$.