You are right.
My weaker demonstration shows how that limit should be, if it exists.
On the other hand, with the previous demonstration not only we know the form of the hypothetical limit, but we are sure that it always exists.
But your sample? what did you mean with post n. #30?
Anyway...
The theorem says that the limit of a sequence product of two other sequences
(with finite limits!) is the product of their limits.
So, in our case ( \lim a_n = A ), if
\lim \sqrt{a_n}
exists finite, than it is
\sqrt{A}
I used the theorem about the limit of products of sequence, as I wrote.
You can use it for sequences with finite limits.
Anyway 1 is not the limit of your sequence, there is no limit for your example.
Do you mean it is a product of two other regular sequences?
We know that
\lim a_n = A
but this is the same as
\lim (\sqrt{a_n})^2 = (\sqrt{A})^2
or
\lim [(\sqrt{a_n}) (\sqrt{a_n})]= (\sqrt{A}) (\sqrt{A})
and, for the theorem about the limit of products of sequence:
\lim [(\sqrt{a_n})...
We know that
\lim a_n = A
but this is the same as
\lim (\sqrt{a_n})^2 = (\sqrt{A})^2
or
\lim [(\sqrt{a_n}) (\sqrt{a_n})]= (\sqrt{A}) (\sqrt{A})
and, for the theorem about the limit of products of sequenze:
\lim [(\sqrt{a_n})...