Where rational functions are undefined

Question

I have another question/comment I'd like a fresh pair of eyes on

The question is "A rational function can have infinitely many x-values at which it is not continuous"

I know since Q(x) in the rational function P(x)/Q(x) can only be factored so many n-th amount of time depending on the degree of Q(x), the statement above is false.

But what about 1/tan(x). Isn't tan(x) undefined at all x-values that are multiple integers of 2(pi) (i.e. x cannot equal 2pi(n))? Do I not assume it's all integers n? In other words, do I just assume that the 1/tan(x) has constraints from [0,2pi)? Hence, excluding the integer n.

Thanks in advance.

Answer 1

A rational function is a quotient of polynomials in which the denominator is not identically zero, i.e. of the form $$R(x) = \frac{P(x)}{Q(x)},$$ $Q \neq 0$. The points at which $R$ is undefined are precisely the zeroes of $Q$. As $Q$ is a polynomial, is has at most $\deg Q$ zeroes.

Answer 2

If you allow polynomials of more than one variable, then it is possible for your function to have more than one root; example:

$P(x,y)= \frac {xy-1}{x^2+y^2-1}$

will be discontinuous everywhere in the unit circle centered at the origin.

Another example, without polynomials. Let $A \subset \mathbb R$ be a connected subset:

$f(x)= \frac {2}{d(x,A)} $ , where $d(x,A):=inf $ {$d(x,a): a $ in $A$} will have uncountably-many zeros (actually, the subset A will be its zero set).