# Can a rational function have infinitely many vertical asymptotes?

My first approach to this problem is to make a rational function that has infinitely many distinct real roots or zeros at distinct values of $$a_1,\dots,a_i$$ for $$n$$ approaching infinity $$\frac{p(x)}{q(x)}=\frac{1}{\prod_{i=1}^n{(x-a_i)}}=\frac{1}{(x-a_1)(x-a_2)\dots(x-a_n)}$$ But I don't think $$q(x)$$ would be considered a polynomial since I think it would have an infinite number of terms, but polynomials can't have infinite terms (I'm not sure if this is actually the case though since I'm saying that $$n\to\infty$$ instead of just $$\infty$$).

• Did you take $p(x)=1$?
– Gary
Sep 22 at 6:51
• You are correct; polynomials by definition always have finitely many terms. And if it has finitely many terms, it cannot have infinitely many zeros (with the zero polynomial being the only exception, but of course you cannot construct a rational function with that in the denominator, while putting it in the numerator causes all so-constructed rational functions to degenerate into the zero polynomial anyway) Sep 22 at 9:38