Limit of a constant sequence

  • Thread starter poutsos.A
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  • #1
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Main Question or Discussion Point

If a sequence {[tex]x_{n}[/tex]} is constant i.e [tex]\ x_{n}=c[/tex] for all nεN how can we prove [tex]limx_{n}[/tex]= c as x goes to infinity??
 

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  • #2
726
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For all epsilon > 0 we have [tex] |x_i - c| = |c - c| = 0 < \epsilon [/tex] where i is any positive integer, thus c is the limit of the sequence. qed
 
  • #3
102
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But the definition of the limit of a sequence says that:

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

[tex]|\ x_{n}-c|<\epsilon[/tex] ,for all n[tex]\geq[/tex] k
 
  • #4
1,074
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But the definition of the limit of a sequence says that:

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

[tex]|\ x_{n}-c|<\epsilon[/tex] ,for all n[tex]\geq[/tex] k
Ok so pick k=1 for all epsilon.
 
  • #5
726
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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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