Problem 4 - Olympiad

Using the Taylor series for e^x: e^x = 1 + x + x^2/2! + x^3/3! + ... = 1 + x + x^2/2 + x^3/6 + ... Then e^x - 1 - x = x^2/2 + x^3/6 + ... Dividing by x^2: (e^x - 1 - x)/x^2 = 1/2 + x/6 + ... As x → 0, this approaches 1/2. Alternatively, using L'Hôpital's rule twice: First application: numerator' = e^x - 1, denominator' = 2x. At x=0: 0/0. Second application: numerator'' = e^x, denominator'' = 2. At x=0: 1/2. The limit is 1/2.