# Continuous with limit

• Jan 23rd 2009, 02:31 PM
Continuous with limit
For $| z | \neq 1$, show that the following limit exists:

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

How to define $f(z)$ when $|z| = 1$ in such a way to make f continuous?

Thanks.
• Jan 23rd 2009, 02:43 PM
vincisonfire
If $z \in \mathbb C$ then you then the range is a unit disk down at -1 on the z-axis. (supposing you work in "kind of" $\mathbb R^3$)
• Jan 24th 2009, 04:25 PM
ThePerfectHacker
Quote:

For $| z | \neq 1$, show that the following limit exists:

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

How to define $f(z)$ when $|z| = 1$ in such a way to make f continuous?

Thanks.

Since $|z|\not = 1$ it means $|z| > 1$ or $|z| < 1$.

If $|z| < 1$ it means $z^n \to 0$ because $|z|^n \to 0$.
Thus, in that case $\lim \frac{z^n-1}{z^n+1} = -1$.

If $|z| > 1$ then $\frac{z^n - 1}{z^n + 1} = \frac{1 - \left( \tfrac{1}{z} \right)^n}{1 + \left( \tfrac{1}{z}\right)^n} \to 1$ because $\left| \frac{1}{z} \right| < 1$.

The function $f$ cannot be extended to a continous function on $\mathbb{C}$. This is because $f$ is not continous on $|z| = 1$. It has one value inside the disk and it has another values outside the disk. Thus, it cannot be continous.