For any given ε > 0, we must find a δ > 0 such that |(2x² - x - 1) - 2| < ε whenever |x - (-1)| = |x + 1| < δ |(2x² - x - 1) - 2| < ε iff |2x² - x - 1 - 2| < ε iff |2x² - x - 3| < ε iff |(2x-3)(x+1)| < ε To find the appropriate δ on the interval (-2,0), we know that |2x-3| < 7 So on the interval (-2,0), |(2x-3)(x+1)| < ε iff Thus whenever δ < ε/7, then |(2x² - x - 1) - 2| < ε thus [PROVED] Edwin