作业帮lim(x->1)(x/(x-1)-1/lnx)这个求极限怎么求
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lim(x->1)(x/(x-1)-1/lnx)这个求极限怎么求
lim(x->1)(x/(x-1)-1/lnx)这个求极限怎么求
lim(x->1)(x/(x-1)-1/lnx)这个求极限怎么求
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
令，t=x-1,则t→0
则原式= lim
=lim ( (t+1)ln(1+t)-t )/(tln(1+t))
=lim ( (t+1)ln(1+t)-t )/t² <等价无穷小代换>
=lim ln(1+t)/2t <洛毕塔法则>
=lim t/2t
=1/2
=lim (xlnx-x+1)/(x-1)lnx
洛必达法则
＝lim(lnx+1-1)/(lnx+(x-1)/x)
=lim (xlnx)/(xlnx+x-1)
继续
=lim (lnx+1)/(lnx+1+1)
=1/2
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