In fact, this is a property of limit, and it is not strictly inequality-preserving.

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.