0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ What is $\operatorname{Arg}z$? $\endgroup$
    – José Carlos Santos
    May 2, 2021 at 9:41
  • 2
    $\begingroup$ Notice that although $\mbox{Arg}z$ has no limit for $z\rightarrow\infty$, it is a limited function taking values in $[0,2\pi)$, while $\ln|z|\rightarrow\infty$ for $z\rightarrow\infty$. So, $$\ln|z|+\mbox{Arg}z=\ln|z|\left(1+\frac{\mbox{Arg}z}{\ln|z|}\right)$$and $\frac{\mbox{Arg}z}{\ln|z|}\rightarrow0$ for $z\rightarrow\infty$, so you remain with the limit $$\lim_{z\rightarrow\infty}\frac{z}{\ln |z|}$$ $\endgroup$
    – Alessandro
    May 2, 2021 at 9:47

1 Answer 1

Reset to default
1
$\begingroup$

$\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}}\gae{Log}\gae{Arg}$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, $\lim_{z\to\infty}\frac z{\Log z}=\infty$.

$\endgroup$
3
  • $\begingroup$ I still don't get why $\frac{\lvert z\rvert}{\ln\lvert z\rvert-2\pi} \rightarrow \infty $ yes $|z|\rightarrow \infty $ but also $ln|z| \rightarrow \infty $ $\endgroup$
    – lupus nox
    May 2, 2021 at 10:26
  • $\begingroup$ @petrosk By l'Hopital $$\lim_{x\to\infty}\frac x{a+\ln x}=\lim_{x\to\infty}\frac{\frac d{dx}[x]}{\frac d{dx}[a+\ln x]}=\lim_{x\to\infty}\frac{1}{1/x}$$ $\endgroup$
    – user239203
    May 2, 2021 at 10:32
  • $\begingroup$ Ok now I see it because every thing is real , we just haven't talked about l'Hopital in my course and i didn't think about it thanks $\endgroup$
    – lupus nox
    May 2, 2021 at 10:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .