好吧.就按这个来：
因f(x)与1/x为无穷小
则lim(x→∞)f(x)=0,lim(x→∞)1/x=0
因f(x)与1/x为等价无穷小
则lim(x→∞)[f(x)/(1/x)]=1
即lim(x→∞)[xf(x)]=1
由此可知f(x)=1/x
所以lim(x→0)[2xf(x)]=2lim(x→0)[x(1/x)]=2lim(x→∞)[xf(x)]=1,可知的是f(x)与1/x为等价无穷小，即f(x)~1/x,不会是f(x)=1/x的这样吧，应该好理因f(x)与1/x为等价无穷小即lim(x→∞)[f(x)/(1/x)]=1即lim(x→∞){[f(x)-(1/x)]/(1/x)}=0于是有f(x)-(1/x)=o[f(x)]即f(x)=1/x+o(f(x))所以lim(x→0)[2xf(x)]=lim(x→0){2x[1/x+o(f(x))]}=lim(x→0)[2+2x*o(f(x))]=2+lim(x→0)[2x*o(f(x))]=2+0=2