1st Putnam 1938

(1)   Find lim_x->inf x²/e^x.

(2)   Find lim_k->0 1/k  ∫₀^k   (1 + sin 2x)^1/x dx.

Solution

(1) Let f(x) = x³e^-x. Then f '(x) = (3x² - x³) e^-x < 0 for x > 3. Hence f(x) < f(3) for x > 3, so x² e^-x < f(3)/x for x > 3. Hence x² e^-x tends to zero.

(2) We use L'Hôpital's rule lim f(x)/g(x) = lim f '(x)/g'(x). Applied to the expression given it gives lim (1 + sin 2x)^1/x. Write (1 + sin 2x)^1/x = exp( 1/x ln(1 + sin 2x) ). So apply the rule again to 1/x ln( 1 + sin 2x) to get 2 cos 2x/(1 + sin 2x) which tends to 2. Hence (1 + sin 2x)^1/x tends to e² and so does the original expression.

1st Putnam 1938

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002