When series {an} and {bn} take a and b as limits respectively, the limit of series {anbn} is a b, and the limit of series {anbn} is ab; When bbn is not equal to 0, the limit of {an/bn} is a/b; When the limits of functions f and g are a and b respectively, the limit of function F B is A B and the limit of function fg is ab; When bg is not equal to 0, the limit of {f/g} is a/b.
The limit problem of sequence is an important part of our study, and the limit theory is also one of the foundations of higher mathematics. As a basic concept of calculus, the limit problem of sequence is of great significance to calculus theory.
The proof methods of four algorithms of sequence limit are as follows:
Theorem: Let {an} and {bn} be convergent sequences, then
( 1)lim(n->； ∞)(an bn)= lim(n-& gt； ∞) a lim (n->; ∞)bn；
(2)lim(n->∞)(an bn)= lim(n-& gt； ∞) a lim (n->; ∞)bn。
If bn≠0 and lim (n->; ∞)bn≠0, then lim (n->; ∞)(an/bn)= lim(n-& gt； ∞)an/lim(n-& gt； ∞)bn。
Certificate: suppose lim (n-> ∞)an=a，lim(n-& gt； ∞)bn=b, then ε >; 0, positive integer n,
When n >; When n, there is | an-a |; ∞)(an+bn)= lim(n-& gt； ∞)an+lim(n-& gt； ∞)bn；
∫an-bn = an+(-bn)，
So lim (n->; ∞)(an-bn)= a-b = lim(n-& gt； ∞)an-lim(n-& gt； ∞)bn。
(2) According to the boundedness theorem, there is a positive number m and there is | bn | < m..
∴|an bn-ab | = | bn(an-a)+a(bn-b)|≤| bn | | an-a |+| a | | bn-b | & lt； (| bn |+| a |)ε& lt； (M+|a|)ε。
∴lim(n->； ∞)(an bn)= lim(n-& gt； ∞) a lim (n->; ∞)bn。
∫an/bn = an 1/bn，so lim(n->； ∞)(an/bn)= lim(n-& gt； ∞)an/lim(n-& gt； ∞)bn。