1st Putnam 1938 Problem A5 (1) Find lim x->inf x 2 /e x . (2) Find lim k->0 1/k ∫ 0 k (1 + sin 2x) 1/x dx.
(1) Let f(x) = x 3 e -x . Then f '(x) = (3x 2 - x 3 ) e -x  3. Hence f(x)  3, so x 2 e -x  3. Hence x 2 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 e 2 and so does the original expression. 1st Putnam 1938 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002