Question

lim[1/(lnx+1)-1/x],x趋于0时的极限，答案是1/2,求详细过程，谢谢

Answer 1

将两个通分：lim[(x-ln(x+1))/xln(x+1)].
用洛必达法则，分子，分母分别求导，求两次后得lim[(1/(x+1)^2)/(ln(x+1)+1/(x+1)^2)]
就可算出等于1/2

Answer 2

原式=lim(x→0)(x-ln(x+1))/[xln(x+1)]=（等价无穷小）lim(x→0)(x-ln(x+1))/x^2=（洛必达法则）lim(x→0)(1-1/(x+1))/(2x)=lim(x→0)(x+1-1)/(2x(x+1))=lim(x→0)1/(2(x+1))=1/2

Answer 3

原式=lim [x-ln(x+1)]/[xln(x+1)]
=lim [1-1/(x+1)/[ln(x+1)+x/(x+1)] 洛必塔法则
=lim 1/[(x+1)+1] 洛必塔法则
=1/2.