Proposition (zero criterion): Let be a polynomial quotient and a a zero of order k for the denominator q, i.e. with . Then the following holds:
f is continuously extendable at a
|a is a zero of at least order k for the numerator p
In this case we can calculate the continuous extension by polynomial division and subsequent cancelling.
Proof: Let A be the domain of f. At first we need to show that a is an accumulation point of A:
As r is continuous we know that r, and with it q as well, is non-zero in whole neighbourhood of a.
Thus q is not the zero polynomial which guarantees that there are only finitely many zeros, for instance . Therefor we find a neighbourhood of a with no zeros of q other than a:
for instance is thus a sequence in converging to a.
Let us now turn to the actual proof:
"": If g denotes the continuous extension of f at a we have for all
which proves a to be a zero of at least order k for p.
"": If the function is a continuous extension of f at a:
g is a polynomial quotient and thus continuous at a