In reply to Ian Wakeling.

Correct. And furthermore, this is a difficult problem to detect and handle. It is very similar to the following example: Use bisection to compute whether the function f(x) = x^2 - epsilon has a root on [-1, 1]. Because bisection uses function evaluations to detect roots, you are likely to miss the fact that the function has a root. There are, of course, different ways to detect roots.