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 $\infty$).
Can a rational function have infinitely many vertical asymptotes
algebra-precalculuspolynomialsrational-functions
Best Answer
The only polynomial function with infinitely many zeros is the null polynomial. But the null polynomial cannot be the denominator of a rational function.