Is $f(z)=\frac{z^{2}-1}{z-1}$ continous at $z=1$?

Question

My teacher said no, but doesn't $f(z)=z+1$ which is continuous?

Answer 1

It's not continuous at $z=1$ because it is not defined there. If you plug $z=1$ in $\frac{z^2-1}{z-1}$, you obtain $\frac{0}{0}$ which is not defined.

You can't divide the numerator by $z-1$ if $z=1$, because then you're dividing by zero!

Answer 2

Basically: The function is not defined at $z=1$ so in particular it cannot be continuous.

But yes: it can be continuously extended by setting $f(1)=2$, see for example What is a continuous extension?

Answer 3

A function $f$ which is continuous at point $t$:
1. should be defined at the point $t$: $f(t)$ exists;
2. should have limit when $x$ tends to $t$: $\lim_{x\to t}f(x)$ exists;
3. $f(t) = \lim_{x\to t}f(x)$ .

In this problem, $f(z)= z+1$ except $z=1$ and not defined at $z=1$.
If we can define $f(z)$ at $1$ , $f(1)=2$ then we can say $f(z)$ is continuous at $z=1$.

Sometimes you can misunderstand that "there exists a limit value" as "a function be continuous at that point".