例 1 设a > 1,证明:limn→∞ = 1.
证:令 = 1 + y n,yn > 0(n=1,2,3,...),应用二项式定理,

, 便得到

于是对于任意给定的ϵ > 0,取N = [],当n > N时,成立

因此limn→∞ = 1
例 2 证明:

证明:令 = 1 + y n,yn > 0(n = 1,2,3,...),应用二项式定理得

即得到

于是,对于任意给定的ϵ > 0,取N = [],
当n > N成立时,成立

因此limn→∞ = 1

定理 0.1 If there are two or more ways to someone will do it.