答案是1/2,先通分,再用洛必达法则求得
lim(x→1) [x/(x-1)-1/lnx]
=lim(x→1) [xlnx-(x-1)]/[(x-1)lnx]
=lim(x→1) (xlnx-x+1)/(xlnx-lnx)
=lim(x→1) (lnx+1-1)/(lnx+1-1/x),洛必达法则
=lim(x→1) lnx/(lnx+1-1/x)
=lim(x→1) (1/x)/(1/x+1/x^2),洛必达法则
=lim(x→1) (1/x)/[(x+1)/x^2]
=lim(x→1) (1/x)*x^2/(x+1)
=lim(x→1) x/(x+1)
=1/(1+1)
=1/2