Originally Posted by

**johnsomeone** If I understand your terminology correctly, I think that using anything more than the definition of the limit is overkill. If f(x)->L as x->c, and L>0, then choose epislon = L/2. This is a standard type of scenario, both the problem and its solution, and so this standard approach is the best way to do it.

By the definition of the limit, when epsilon = L/2 (note it's positive), there's some positive delta that defines a corresponding deleted open neighborhood U of c

( U = [a,b] intersect (c-delta, c+delta) intersect {c}-complement )

such that for all x in U, abs(f(x)-L) < L/2. Thus for all x in U, -L/2 < f(x)-L, and thus f(x) > L/2. Since L/2 > 0, have shown that there exists a deleted open neighborhood U of c such that for all x in U, f(x) is positive.