HMMT 二月 1998 · CALC 赛 · 第 5 题

英文解析

5. Problem: Evaluate lim x .
   x → 1
   sin(1 − x )xxsin(1 − x )
   ln xln x
   Solution: Rewrite the expression to evaluate as e . Then we must evaluate lim e .
   x → 1
   ( )
   sin(1 − x )xxlim ln x = lim ln x . Because direct calculation of the limit results in indeterminatex → 1 x → 1
   sin(1 − x )
   ( )
   1 xform ( · 0), we can use L’Hopital’s rule to evaluate the limit. By L’Hopital’s rule, lim ln x =
   x → 10
   sin(1 − x )
   ln x + 1
   lim . This limit is simply -1.
   x → 1 − cos(1 − x )
   x 1
   sin(1 − x )
   ln x − 1 1
   Hence lim e = e = .
   x → 1 e