Choose δ ∈ (0,1) such that 1 - δ < x < 1 implies 3/M < 2x^{2}- 3x + 1 < 0; i.e., M/3 > 1/(2x^{2}- 3x + 1). Notice that 0 < x < 1 also implies 2 < x + 2 < 3. It follows that f(x) = (x+2)/(2x^{2}- 3x + 1) < M for all 1 - δ < x < 1.

I do not understand the last statement because it seems to say that if 0 < a < b and d < c < 0, then ad < bc. Is this true?