HMMT 二月 2008 · CALC 赛 · 第 7 题

英文解析

7. [ 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 −
   lim t + 1 + − 1 − 2 = lim t 1 + t + 1 − t − 2 .3
   t → 0 t → 0
   t t t
   √ √
   3 3
   We need the degree of the first nonzero term in the Mac laurin expansion of 1 + t + 1 − t − 2. Wehave
   √ √
   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
   1 59
   follows that − p − = − 2. So p = .
   3 3
   ∫
   ln 2
   3 x 2 x
   2 e + e − 1 T