其他答案

2011-12-03 19:42:49

• 原式=lim(x^2-sin^2 x)/(x^2*sin^2 x)
  =lim(2x-2sinxcosx)/(2x*sin^2 x+x^2*2sinxcosx)
  =lim[x-(1/2)sin2x]/[x*sin^2 x+x^2*(1/2)sin2x]
  =lim(1-cos2x)/(sin^2 x+x*sin2x+x*sin2x+x^2*cos2x)
  =lim(1-cos2x)/(sin^2 x+2x*sin2x+x^2*cos2x)
  =lim[1-(1-2sin^2 x)]/(sin^2 x+2x*sin2x+x^2*cosx)
  =lim2sin^2 x/(sin^2 x+2x*sin2x)+x^2*cosx)
  =2/(1+4+1)
  =1/3.

  全部
  

  T***

  2011-12-03 19:42:49

  43 2 评论

  分享
  提交评论