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)]=2
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)]=2