limx->1求极限(x/x-1 - 1/lnx)

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/15 17:46:17
limx->1求极限(x/x-1-1/lnx)limx->1求极限(x/x-1-1/lnx)limx->1求极限(x/x-1-1/lnx)lim(x->1)[x/(x-1)-1/lnx]=lim(x-

limx->1求极限(x/x-1 - 1/lnx)
limx->1求极限(x/x-1 - 1/lnx)

limx->1求极限(x/x-1 - 1/lnx)
lim(x->1)[ x/(x-1) - 1/lnx ]
=lim(x->1) [xlnx-(x-1)]/[(x-1)lnx] (0/0)
= lim(x->1) [ (1+lnx-1) / (lnx + (x-1)/x) ]
= lim(x->1) [ xlnx/(xlnx+(x-1) ] (0/0)
= lim(x->1) [ (lnx+1)/(x+1+1) ]
=1/2

limx->1 (x/x-1 - 1/lnx)
=limx->1 (xlnx-x+1)/[(x-1)lnx] (0/0)
=limx->1 lnx/[lnx+(x-1)/x]
=limx->1 xlnx/[xlnx+(x-1)] (0/0)
=limx->1 (lnx+1)/[lnx+2]
=1/2