HMMT 二月 2003 · CALC 赛 · 第 4 题

英文解析

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
   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