Question

lim(x->1)(x/(x-1)-1/lnx)这个求极限怎么求

Answer 1

取极限就不写了,除最后一行每步之前加上
limx→1就可以了

原式=x/(x-1)-1/lnx=1+1/(x-1)-1/lnx=1+(lnx-x+1)/lnx(x-1)
(分子分母分别求导)

=1+(1/x-1)/[lnx+(x-1)/x]=1+(1-x)/(xlnx+x-1)
(分子分母分别求导)

=1+(-1)/(lnx+x×1/x+1)=1-1/(2+lnx)

=1-1/(2+ln1)=1-1/2=1/2

Answer 2

lim(x->1)(x/(x-1)-1/lnx)
=lim(x->1)(xlnx-x-1)/[(x-1)lnx]（这是0/0型，运用洛必达法则）
=lim(x->1)(lnx+1-1)/[lnx+(x-1)/x]
=lim(x->1)(xlnx)/[xlnx+(x-1)]
(这是0/0型，运用洛必达法则）
=lim(x->1)(lnx+1)/[lnx+1+1]
=1/2