Question

用分析定义证明lim(x->2)x^2=4

才学高数。。很多证明都不是很明白
用分析定义证明lim(x->2)x^2=4
答案上写：
任意ε>0,存在δ=min(1,ε/5),当0<│x-2│<δ时总有│x^2-4│<ε
。不太理解啊。帮帮忙啦~~

Answer 1

证明：任给正数ε>0,要使│x^2-4│<ε成立,只要(4-ε)^(1/2)<x<(4+ε)^(1/2)，
则-[2-(4-ε)^(1/2)]<x-2<(4+ε)^(1/2)-2
因为2-(4-ε)^(1/2)=ε/[2+(4-ε)^(1/2)]>(4+ε)^(1/2)-2=ε/[2+(4+ε)^(1/2)]
故取δ=(4+ε)^(1/2)-2,当0<│x-2│<δ时总有│x^2-4│<ε
这就证明了lim(x->2)x^2=4