Question

当x趋于0时lim(1/x-1/tanx) 的极限,咋做 详细点

Answer 1

lim(1/x-1/tanx)=lim (1/x-cosx/sinx)
简单的说
当x->0时,cosx->1,sinx->x
所以，应该猜到极限是0。
lim(1/x-1/tanx)=lim (1/x-cosx/sinx)
=lim (sinx-xcosx)/(xsinx)
上下求导
=lim (cosx -cosx+xsinx)/(sinx+xcosx)
=lim (xsinx)/(sinx+xcosx)
=lim sinx/((sinx)/x+cosx)
lim sinx =0, lim (sinx)/x=1 lim cosx=1
故lim(1/x-1/tanx)=lim (1/x-cosx/sinx)=lim sinx/((sinx)/x+cosx)=0

追问

我其实想问是1/x2-1/tanx2，就是刚才那个x都变成x的平方。答案似乎是三分之二或者二分之三 ，你会不

追答

答案是0
令t=x^2,x->0 => t->0
故 lim 1/t-1/tant=0
即lim(1/x^2-1/tan(x^2))=0
或按照类似过程求解
lim(1/x^2-1/tan(x^2))=lim (1/x^2-cos(x^2)/sin(x^2))
=lim (sin(x^2)-x^2cos(x^2))/(x^2sin(x^2))
上下求导
=lim (2x*cos(x^2)-2xcos(x^2)+2x^3*sin(x^2))/(2x*sin(x^2)+2x^3*cos(x^2))
=lim 2x^3*sin(x^2)/(2x*sin(x^2)+2x^3*cos(x^2))
=lim sin(x^2)/(sin(x^2)/x^2+cos(x^2))
=0
但是楼主的题目如果是
lim(1/x^2-1/(tanx)^2)
那么答案是2/3
lim(1/x^2-1/(tanx)^2)=
=lim (1/x^2-(cosx)^2/(sinx)^2)
=lim [(sinx)^2-x^2(cosx)^2]/[x^2(sinx)^2]
上下求导
=lim [2sinx*cosx-2x(cosx)^2+2x^2*cosx*sinx]/[2x(sinx)^2+2x^2sinx*cosx]
=lim [sin2x-x(cos2x+1)+x^2*sin2x]/[x(1-cos2x)+x^2*sin2x]
=lim [2cos2x -cos2x-1+2x*sin2x+2x*sin2x+2x^2*cos2x]/[1-cos2x+2x*sin2x+2x*sin2x+2x^2*cos2x]
=lim [cos2x-1+4x*sin2x+2x^2*cos2x]/[1-cos2x+4x*sin2x+2x^2*cos2x]
=lim [-2sin2x+4sin2x+8x*cos2x+4x*cos2x-4x^2*sin2x]/[2sin2x+4sin2x+8x*cos2x+4x*cos2x-4x^2*sin2x]
=lim [2sin2x+12x*cos2x-4x^2*sin2x]/[6sin2x+12x*cos2x-4x^2*sin2x]
上下同时除以2x
=lim [2(sin2x)/2x+6*cos2x-2x*sin2x]/[6(sin2x)/2x+6cos2x-2x*sin2x]
= [2*1+6-0]/[6*1+6-0]
=8/12=2/3