当然可以：先分母有理化
ln(2+x)/((1+2x)^(1/3)+1)
=ln(2+x)*((1+2x)^(2/3)+(1+2x)^(1/3)+1)/((1+2x+1)
=(1/2)[ln(2+x)/(x+1)]*((1+2x)^(2/3)+(1+2x)^(1/3)+1)
lim((1+2x)^(2/3)+(1+2x)^(1/3)+1)=1
limln(2+x)/(x+1)]=limln(1+x+1)^(1/(x+1))=lne=1
所以：原极限=1/2