Proving the Limit of a Constant Sequence

poutsos.A
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If a sequence {x_{n}} is constant i.e \ x_{n}=c for all nεN how can we prove limx_{n}= c as x goes to infinity??
 
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For all epsilon > 0 we have |x_i - c| = |c - c| = 0 < \epsilon where i is any positive integer, thus c is the limit of the sequence. qed
 
But the definition of the limit of a sequence says that:

lim\ x_{n} = c iff for all ε>0 there exists a k belonging to the natural Nos N SUCH that :

|\ x_{n}-c|<\epsilon ,for all n\geq k
 
poutsos.A said:
But the definition of the limit of a sequence says that:

lim\ x_{n} = c iff for all ε>0 there exists a k belonging to the natural Nos N SUCH that :

|\ x_{n}-c|<\epsilon ,for all n\geq k

Ok so pick k=1 for all epsilon.
 
I don't see the problem. Since the sequence is a constant sequence, each x_i is equal to each other, so as long as we take i >= N = 1, we will have |c-c| = 0 < epsilon, proving that c is the limit of the sequence. Even if we took N = 2389472389432, any i after that is still equal to c.
 
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