Note: You may want to wait for a more senior member to validate this answer.
My interpretation of your question
"If
with
, and if
is not a limit point of
then does
make sense?"
Answer to that question: No. For
to exist there must be a point
such that for every
there exists a
such that
. But now suppose that
is not a limit point of
then we can pick an open neighborhood of
lets call it
such that there does not exist an element of
in that neighborhood. Now choose an value of
that requires
and we run into a contradiction.
EDIT: I am sorry, I completely for some reason disregarded the
part. This can be made slightly simpler
By the defintion of limits in
for
to exist there must exist a point
such that for every
there exsits a
such that
. But now suppose that
is not a limit point of
then there exists a neighborhood
such that there does not exist an element of
within that neighborhood. Now if we choose an
that requires
we run into a contradiction.