还是老样子,极限的定义,无限分有限+无限
lim(x(n+1)-x(n))/(y(n+1)-yn)存在
设lim(x(n+1)-x(n))/(y(n+1)-yn)=a
对于任意e>0,存在N使得,对n>N有
|(x(n+1)-x(n))/(y(n+1)-yn)-a|那么对于n>N,有
a-e<(x(n+1)-x(n))/(y(n+1)-yn)(a-e)(y(n+1)-yn)那么
(a-e)(y(N+2)-y(N+1))(a-e)(y(N+3)-y(N+2))...
(a-e)(y(n+1)-yn)相加有
(a-e)(y(n+1)-y(N+1))故
|(x(n+1)-x(N+1))/(y(n+1)-y(N+1))-a|现在要转化xn/yn为含有上式的形式,并证明其极限
xn/yn - a=(xn-x(N+1))/(yn-y(N+1))* (yn-y(N+1))/yn+(x(N+1)-a*y(N+1))/yn 凑出上式
|xn/yn - a|<=e|1-y(N+1)/yn|+|(x(N+1)-a*y(N+1))/yn|<=e+|(x(N+1)-a*y(N+1))/yn|
存在N'>N使得对n>N'有
|(x(N+1)-a*y(N+1))/yn|那么对于任意e>0
有|xn/yn - a|<2e
那么lim xn/yn=lim(x(n+1)-x(n))/(y(n+1)-yn)