求极限 答案是e的-1次

发布时间:2021-02-20 16:58:23

求极限 答案是e的-1次

网友回答

设:y = ((3-e^x)/(x+2))^(1/sin(x))
lny=1/sin(x)*ln((3-e^x)/(x+2));
现求lim(1/sin(x)*ln((3-e^x)/(x+2)))当x->0时极限;对上式进行求导得:
(-e^x*ln(e)/(x+2)-(3-e^x)/(x+2)^2)/cos(x)
把x=0,代入得lny=-1;
故y=e^-1
======以下答案可供参考======
供参考答案1:
  利用等价无穷小
   ln(1+x) ~ x ,sinx ~ x (x→0),
极限   lim(x→0)(1/sinx)*ln[(3-e^x)/(x+2)]
  = lim(x→0)(1/sinx)*ln[1+(1-x-e^x)/(x+2)]
  = lim(x→0)(1/sinx)*[(1-x-e^x)/(x+2)]
  = lim(x→0)[1/(x+2)]*lim(x→0)[(1-x-e^x)/x]
  = (1/2)*lim(x→0)[(1-x-e^x)/x] (0/0)
  = (1/2)*lim(x→0)[(-1-e^x)/1]
  = (1/2)*(-2)
  = -1,
所以  g.e. = e^(-1)。
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