How to show that the limit exists for a complex function?

Question

When $\lvert z\rvert \ne 1$, it is necessary to show the existence of the limit of the function

$$f(z) = \lim\limits_{n\to\infty}\frac{z^n-1}{z^n+1}$$

Answer 1

First, when $|z|<1$,$$\lim_{n\to\infty}z^n=0.$$ So case one: when $|z|<1$, $$\lim\limits_{n\to\infty}\frac{z^n-1}{z^n+1}=\frac{\lim\limits_{n\to\infty}z^n-1}{\lim\limits_{n\to\infty}z^n+1}=\frac{0-1}{0+1}=-1.$$ Case two： when $|z|>1$, then $\left|\frac{1}{z}\right|<1$,$\lim\limits_{n\to\infty}\frac{1}{z^n}=0$. $$\lim\limits_{n\to\infty}\frac{z^n-1}{z^n+1}=\lim\limits_{n\to\infty}\frac{1-\frac{1}{z^n}}{1+\frac{1}{z^n}}=\frac{1-\lim\limits_{n\to\infty}\frac{1}{z^n}}{1+\lim\limits_{n\to\infty}\frac{1}{z^n}}=\frac{1-0}{1+0}=1.$$

Answer 2

Consider that $$f(z) = \lim_{n\to\infty}\frac{z^n-1}{z^n+1}=1-2\lim_{n\to\infty}\frac{1}{z^n+1}$$ if $|z|>1$ then $|z|^n\to\infty$ and $$|\frac{1}{z^n+1}|\leq\frac{1}{|z|^n-1}\to0$$ then limit of $f$ is $\color{blue}{1}$.

If $|z|<1$ then $|z|^n\to0$ and $$\frac{|z^n|}{|z|^n+1}\leq\left|\frac{z^n}{z^n+1}\right|\leq\frac{|z^n|}{1-|z|^n}$$ shows this limit is $0$, Therfore when $|z|<1$, $$f(z) = \lim_{n\to\infty}\frac{z^n-1}{z^n+1}=2\lim_{n\to\infty}\frac{z^n}{z^n+1}-1$$ and the limit is $\color{blue}{-1}$.