当x趋于0时lim(1/x-1/tanx) 的极限,咋做

当x趋于0时lim(1/x-1/tanx) 的极限,咋做

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
再问： 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
再答： 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