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

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?

• What is $\operatorname{Arg}z$? May 2, 2021 at 9:41
• 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 $$\lim_{z\rightarrow\infty}\frac{z}{\ln |z|}$$ May 2, 2021 at 9:47

$$\newcommand{gae}{\newcommand{#1}{\operatorname{#1}}}\gae{Log}\gae{Arg}$$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, $$\lim_{z\to\infty}\frac z{\Log z}=\infty$$.
• I still don't get why $\frac{\lvert z\rvert}{\ln\lvert z\rvert-2\pi} \rightarrow \infty$ yes $|z|\rightarrow \infty$ but also $ln|z| \rightarrow \infty$ May 2, 2021 at 10:26
• @petrosk By l'Hopital $$\lim_{x\to\infty}\frac x{a+\ln x}=\lim_{x\to\infty}\frac{\frac d{dx}[x]}{\frac d{dx}[a+\ln x]}=\lim_{x\to\infty}\frac{1}{1/x}$$