Show that the function $f$ is continuous at the given point

$$f(z) =\begin{cases} \dfrac{z^3 - 1}{z-1} & \text{if } |z| \neq 1 \\[6pt] 3 & \text{if } |z| = 1 \\ \end{cases}$$ $z_0 = 1$

I have the following definition for the Limit of a Complex Function:

Suppose that a complex function $f$ is defined in a deleted neighborhood of $z_0$ and suppose that $L$ is a complex number. The limit of $f$ as $z$ tends to $z_0$ exists and is equal to $L$, written as $\lim\limits_{z \to z_0} f(z) = L$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $\vert f(z) - L \vert < \epsilon $ whenever $0 < \vert z - z_0 \vert < \delta$.

My intuition is to start with $\vert f(z) - L \vert < \epsilon$ and manipulate this until the expression on the left looks like $\vert z-z_0 \vert$. From there, I figured I would have identified an appropriate value for my $\delta$ and used it to work forward for my proof. However, I got a little stuck with the manipulation. I can see that for $z = z_0 = 1$ that $\vert z_0 \vert = 1$ and so $f(z_0) = 3$ immediately. When I work in the neighborhood of $z_0$ for $\vert z \vert \neq 1$ that I do approach $3$. I've decided to try an epsilon/delta proof for the condition that $vert z \vert \neq 1$. Here's what I've done so far:

$$ \begin{align} f(z) & = \frac{z^3 - 1}{z-1} \\[6pt] & = \frac{z^3 - 1^3}{z-1} \\[6pt] & = \frac{(z-1)(z^2 + z + 1^2)}{z-1} \\[6pt] & = z^2 + z + 1 \end{align} $$

From here, I can plug this alternate form of $f(z)$ into $\vert f(z) - L \vert < \epsilon$ and continue to manipulate:

$$ \begin{align} \varepsilon & > \vert (z^2 + z + 1) - (3) \vert \\[6pt] & = \vert z^2 + z - 2 \vert \\[6pt] & = \vert (z-1)(z+2) \vert \end{align} $$

This is where I'm stuck. I don't think that I can just divide $\epsilon$ by $\vert z+2 \vert$ because that would make it depend on the variable $z$ which defeats setting a fixed value for $\delta$. What am I supposed to do here?


That's good. Now just observe that you can take $\delta<1$, so $|z+2|<4$ and therefore, for $|z-1|<\delta=\min(\varepsilon/4,1)$ you have $$ |(z^2+z+1)-3|=|z-1|\,|z+2|<\frac{\varepsilon}{4}\cdot 4=\varepsilon $$

Note that, if $|z-1|<1$, you have $$ |z+2|=|(z-1)+3|\le |z-1|+3<4 $$

On the other hand, the function $g(z)=z^2+z+1$ is everywhere continuous and coincides with $f$ on $\mathbb{C}\setminus\{1\}$. Therefore $$ \lim_{z\to 1}f(z)=\lim_{z\to 1}g(z)=g(1)=3 $$ by definition of limit and continuity.

  • $\begingroup$ Thank you so much for your help. That was elegant. $\endgroup$ – bloodtypebpos Feb 21 '16 at 17:36
  • $\begingroup$ @bloodtypebpos More generally, if $g$ is continuous on some open set $U$ and differentiable at $z_0\in U$, the function $f(z)=(g(z)-g(z_0))/(z-z_0)$, for $z\ne z_0$, and $f(z_0)=g'(z_0)$ is continuous at $z_0$ (and so in the whole set $U$). $\endgroup$ – egreg Feb 21 '16 at 17:40

Observe that $$\vert (z-1)(z-1+3) \vert \leq \vert (z-1)^2 \vert + 3\vert (z-1) \vert \leq \delta^2+3\delta.$$ Therefore consider the quadratic equation $$\delta^2+3\delta-\epsilon=0$$ Solving this should give you a positive $\delta$ in terms of $\epsilon$ which you desire.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.