1. Show that lim(x→c) 1/g(x) = 1/L2, for L2 ≠ 0
2. Make |1/g(x) - 1/L2| as small as desired
3. Numerator of fraction on right-hand side of (1.8) can be made as small as needed
4. Consider what happens to the denominator
5. For ε2 = |L2|/2, |g(x) - L2| < |L2|/2
6. By triangle inequality, |L2| = |L2 - g(x) + g(x)| ≤ |L2 - g(x)| + |g(x)| < |L2|/2 + |g(x)|
7. Subtracting |L2|/2 from both sides gives |L2|/2 < |g(x)|
9. From (1.8), for 0 < |x - c| < δ2, |1/g(x) - 1/L2| < 2|L2 - g(x)|/L2^2
10. Given ε > 0, ∃ δ3 > 0 such that |L2 - g(x)| < εL2^2/2, whenever 0 < |x - c| < δ3
11. For δ = min{δ2, δ3}, |1/g(x) - 1/L2| < ε, whenever 0 < |x - c| < δ