lim(x→-1）ln(2+x)=0lim(x→-1）(1+2x)^(1/3) +1=0罗比达法则lim(x→-1）ln(2+x)/[(1+2x)^(1/3)+1]=lim(x→-1）[ln(2+x)]' / [(1+2x)^(1/3)+1]'=lim(x→-1）[1/(2+x)] /[(2/3) (1+2x)^(-2/3)]=1/(2/3)=3/2...太感谢了~~~~~