数学分析 数列 极限 证明
证明:若(1)y(n+1)>y(n)(2)limyn->∞(3)lim(x(n+1)-x(n))/(y(n+1)-yn)存在那么limxn/yn=lim(x(n+1)-x...
证明:若
(1)y(n+1)>y(n)
(2)lim yn->∞
(3)lim(x(n+1)-x(n))/(y(n+1)-yn)存在
那么lim xn/yn=lim(x(n+1)-x(n))/(y(n+1)-yn) 展开
(1)y(n+1)>y(n)
(2)lim yn->∞
(3)lim(x(n+1)-x(n))/(y(n+1)-yn)存在
那么lim xn/yn=lim(x(n+1)-x(n))/(y(n+1)-yn) 展开
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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|<e
那么对于n>N,有
a-e<(x(n+1)-x(n))/(y(n+1)-yn)<a+e
(a-e)(y(n+1)-yn)<x(n+1)-xn<(a+e)(y(n+1)-yn)
那么
(a-e)(y(N+2)-y(N+1))<x(N+2)-x(N+1)<(a+e)(y(N+2)-y(N+1))
(a-e)(y(N+3)-y(N+2))<x(N+3)-x(N+2)<(a+e)(y(N+3)-y(N+2))
...
(a-e)(y(n+1)-yn)<x(n+1)-xn<(a+e)(y(n+1)-yn)
相加有
(a-e)(y(n+1)-y(N+1))<x(n+1)-x(N+1)<(a+e)(y(n+1)-y(N+1))
故
|(x(n+1)-x(N+1))/(y(n+1)-y(N+1))-a|<e
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|<e
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)
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|<e
那么对于n>N,有
a-e<(x(n+1)-x(n))/(y(n+1)-yn)<a+e
(a-e)(y(n+1)-yn)<x(n+1)-xn<(a+e)(y(n+1)-yn)
那么
(a-e)(y(N+2)-y(N+1))<x(N+2)-x(N+1)<(a+e)(y(N+2)-y(N+1))
(a-e)(y(N+3)-y(N+2))<x(N+3)-x(N+2)<(a+e)(y(N+3)-y(N+2))
...
(a-e)(y(n+1)-yn)<x(n+1)-xn<(a+e)(y(n+1)-yn)
相加有
(a-e)(y(n+1)-y(N+1))<x(n+1)-x(N+1)<(a+e)(y(n+1)-y(N+1))
故
|(x(n+1)-x(N+1))/(y(n+1)-y(N+1))-a|<e
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|<e
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)
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