求极限 答案是e的-1次

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 01:28:08

求极限 答案是e的-1次
求极限 答案是e的-1次

求极限 答案是e的-1次
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

  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...

全部展开

  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)。

收起