Question

数学极限相关。关于两个重要极限的运用

Answer 1

1、原式=lim(sinx-sinx/cosx)/x^3=sinx(cosx-1)/(cosxx^3)，因1-cosx=2(sinx/2)^2，
原式=lim-[2sinx(sinx/2)^2/(cosxx^3)=-2•1/4=-1/2
2、设3x=tany，则原式=lim(3y/tany)=lim3(ycosy/siny)=3

追问

果然第一个答案错了

课本上的答案

Answer 2

(1) lim(x→0) (sinx-tanx)/x³
=lim(x→0) (sinx/x)*[(1-1/cosx)/x²]
=lim(x→0) (sinx/x)*[(cosx-1)/x²cosx]
=lim(x→0) (sinx/x)*[-2sin²(x/2)/x²cosx]
=lim(x→0) (sinx/x)*[-sin²(x/2)/2(x/2)²cosx]
=(-1/2)*lim(x→0) (sinx/x)*[sin(x/2)/(x/2)]²*(1/cosx)
=(-1/2)*lim(x→0) 1/cosx
=-1/2

(2)令t=arctan3x, 则x=tant/3, 且x→0时t→0
lim(x→0) arctan3x/x
=lim(t→0) 3t/tant
=lim(t→0) 3tcost/sint
=lim(t→0) 3cost
=3