HMMT 二月 2003 · CALC 赛 · 第 4 题

HMMT February 2003 — CALC Round — Problem 4

The sequence of real numbers x , x , x , . . . satisfies lim ( x + x ) = 315 and 1 2 3 n →∞ 2 n 2 n +1 lim ( x + x ) = 2003. Evaluate lim ( x /x ). n →∞ 2 n 2 n − 1 n →∞ 2 n 2 n +1 4

解析

The sequence of real numbers x , x , x , . . . satisfies lim ( x + x ) = 315 and 1 2 3 n →∞ 2 n 2 n +1 lim ( x + x ) = 2003. Evaluate lim ( x /x ). n →∞ 2 n 2 n − 1 n →∞ 2 n 2 n +1 Solution: − 1 We have lim ( x − x ) = lim [( x + x ) − ( x + x )] = 315 − 2003 = n →∞ 2 n +1 2 n − 1 n →∞ 2 n 2 n +1 2 n 2 n − 1 − 1688; it follows that x → −∞ as n → ∞ . Then 2 n +1 x x + x 2 n 2 n 2 n +1 lim = lim − 1 = − 1 , n →∞ n →∞ x x 2 n +1 2 n +1 since x + x → 315 while x → −∞ . 2 n 2 n +1 2 n +1 4