It is also easy to prove:
For the convergent sequence {an} (convergent to a), if an < C and c is constant, then a ≤ c.
Because lim an=a, by definition,
Any ε > 0, with n1>; 0, when n>N 1, where | an-a | < ε/2.
So there is a -ε/2.
At the same time, according to the definition, lim c = C.
For ε > above; 0, N2 > exists; 0, when n & gtN2, use | c-c | < ε/2.
Therefore, there is C.
There is a<c, namely: a-ε/2.
That is to say,
For any ε > 0, a< has one.
The above proof uses a proposition:
Let a and b be two real constants, then the necessary and sufficient conditions for a≤b are: arbitrary ε >; 0, a< has one.
It's easy to prove by reduction to absurdity ~ ~ ~
In fact, this is not difficult to understand, because the limit itself has the property of breaking strict inequalities.
For example, for any n>0, there must be 1/n > 0, but after taking the limit, lim 1/n=0=lim 0=0.
So the strict inequality was broken ~ ~
If you understand, please ask questions.