Question

$f: \overline{B(0,1)} \to \mathbb{C}$ $f(z) = \begin{cases} 3z^2 & |z| < 1\\ 3 & |z| = 1 \end{cases}$ Prove $\lim_{z \to z_0} f(z)$ does not exist for any $z_0$ with $|z_0| = 1$ $\quad z_0 \neq \pm 1$ using $\epsilon, \delta$ definition.

          $f: \overline{B(0,1)} \to \mathbb{C}$
$f(z) = \begin{cases} 3z^2 & |z| < 1\\ 3 & |z| = 1 \end{cases}$
Prove $\lim_{z \to z_0} f(z)$ does not exist
for any $z_0$ with $|z_0| = 1$
$\quad z_0 \neq \pm 1$
using $\epsilon, \delta$ definition.

f: B(0,1)→ℂ
f(z) = 3z^2 |z| < 1
3 |z| = 1
Prove limz → z0 f(z) does not exist
for any z0 with |z0| = 1
z0 ≠± 1
using ϵ, δ definition.

Added by Gabriel V.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Supreeta N

12/01/2023

Step 1

Step 1: Let $z_0$ be any complex number such that $|z_0| = 1$ and $z_0 \neq \pm 1$. Show more…

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$J: B(0,1) \rightarrow \mathbb{C}$. $f(z) = \begin{cases} 3z^2 & |z| < 1 \\ 3 & |z| = 1. \end{cases}$ Prove $\lim_{z \to z_0} f(z)$ does not exist for any $z_0$ with $|z_0| = 1$ and $z_0 \neq \pm 1$ using $\epsilon, \delta$ definition.