I am trying to solve a tricky textbook problem that requires the use of the epsilon-delta definition of a limit. However, with my elementary understanding of how to answer these kinds of problems, the solution on the back of the book begins its solution by stating something along these lines:
Since f(point of interest) can't be 0 [in this question's scenario], and the function is continuous on the point of interest, then you can say:
$$\varepsilon = \frac{|f(\text{point of interest})|}{2} > 0$$
How is this possible? Correct me if I'm wrong, but I thought epsilon was only calculable through direct manipulation of delta. How can you make this deduction without delta, just from knowing that the function at a point of interest can't be $0$?
Any help please? Thank you.
PS: The only information given for δ is it's non-negative ($δ >0$). That is it.