Approximate e^0.05 and e^x for small x. What's the rule?

How to answer

For small x, e^x ≈ 1 + x + x²/2. e^0.05 ≈ 1.05125 (true 1.05127). First-order e^x≈1+x carries ~x²/2 error. Inversely ln(1+x) ≈ x − x²/2, so ln(1.05) ≈ 0.04875 (true 0.04879). The gap between continuous and simple 5% interest is ~12.7 bps.

Key idea: Using e^x≈1+x for large x (e^0.5≈1.5 is ~9% low; true 1.6487). Add the quadratic term beyond x≈0.1.