是求x[ln(x+1)-ln(x)]的极限吧?
lim(x->∞)x[ln(x+1)-ln(x)]
= lim(x->∞)ln((x+1)/x)/(1/x) (0/0型罗比塔法则)
= lim(x->∞)(x/(x+1))*(-1/x^2)/(-1/x^2)
= lim(x->∞)x/(x+1)
= 1