计算题,计算lim(x趋向1) [(x/x-1)-(1/lnx)]=发错位置……估计是难有人回答了

答过了
lim [(x/x-1)-(1/lnx)]
=lim [(xlnx-x+1)/((x-1)lnx)] ------(通分)
=lim[lnx/((x-1)/x+lnx)]----(0/0型用洛贝塔法则分子分母求导)
=lim[lnx/(1-1/x+lnx)]
=lim[(1/x)/(1/x²+1/x)]-----(仍然是0/0型继续用洛贝塔法则分子分母求导)
=1/2------(代入x=1)
所以lim(x趋向1) [(x/x-1)-(1/lnx)]=1/2

lim(x->1)x/(x-1)-1/lnx
=lim(x->1)[xlnx-x+1]/(x-1)lnx
应用洛必达法则
=L'=lim(x->1)(lnx+1-1)/[lnx+1-1/x]
=lim(x->1)(1/x)/(1/x+1/x^2)
=1/2