HMMT 二月 2008 · CALC 赛 · 第 7 题

HMMT February 2008 — CALC Round — Problem 7

[ 5 ] Find p so that lim x x + 1 + x − 1 − 2 x is some non-zero real number. x →∞ ∫ ln 2 3 x 2 x 2 e + e − 1 T

解析

[ 5 ] Find p so that lim x x + 1 + x − 1 − 2 x is some non-zero real number. x →∞ 5 1 Answer: Make the substitution t = . Then the limit equals to 3 x ( ) √ √ √ ( √ √ ) 1 1 1 1 3 3 3 3 3 − p − p − 3 lim t + 1 + − 1 − 2 = lim t 1 + t + 1 − t − 2 . t → 0 t → 0 t t t √ √ 3 3 We need the degree of the first nonzero term in the MacLaurin expansion of 1 + t + 1 − t − 2. We have √ √ 1 1 1 1 3 3 2 2 2 2 1 + t = 1 + t − t + o ( t ) , 1 − t = 1 − t − t + o ( t ) . 3 9 3 9 √ √ 3 3 2 2 2 It follows that 1 + t + 1 − t − 2 = − t + o ( t ). By consider the degree of the leading term, it 9 1 5 follows that − p − = − 2. So p = . 3 3 ∫ ln 2 3 x 2 x 2 e + e − 1 T