Let {X_n} be a sequence of real numbers that is bounded above by M and such that X_n-->x Prove that x 0 then there exist N>=1 s.t. n>=N. then |x_n - x| =N x-m<x_n<x. then -M<X+X_n then x_n+x<M

Let {X_n} be a sequence of real numbers that is bounded above by M and such that X_n-->x Prove that x<=M.

This is what I have, but I believe it is wrong:
Let M>0 then there exist N>=1 s.t. n>=N. then |x_n - x|<M. Since {x_n} is bounded above by the given {x_n} has a least upper bound(sup). Let x=Lub{x}. Then x-M<x. then for n>=N x-m<x_n<x. then -M<X+X_n then x_n+x<M

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