Proving the Limit of a Constant Sequence

[poutsos.A]

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??

 

[JG89]

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

 

[poutsos.A]

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

 

[d_leet]

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.

 

[JG89]

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.