Question

求证大学微积分的数列极限题

利用数列极限的定义证明：数列Xn=(n+2/n^2-2)sin n 的极限为0

Answer 1

|Xn-0|=|(n+2)/(n^2-2)||sin n|<=|(n+2)/(n^2-2)|
对任意eps
令N=max(2,4/eps)
对于任意n>N
有
|Xn-0|<=|(n+2)/(n^2-2)|
<=2|(n+2)/n^2| (因为(n^2)/2=<n^2-2, 合一起得n^2>=4)
<=2(2n)/n^2(因为n+2<=2n,2<=n,因为n>N>=2)
<=4/n
<=4/(4/eps)
<=eps