>When f(0)≠0, we can state decisively that there is no such Q This is false. This relies on an assumption not stated - that f is continuous. For a simple coun

>When f(0)≠0, we can state decisively that there is no such Q This is false. This relies on an assumption not stated - that f is continuous. For a simple counterexample: f(x) = 1, x=0 f(x) = x, x!=0 g(x) = x Limit of the ratio as x approaches 0 is 1.