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 (x...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2/3lim(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