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Def: A sequence $X = (x_n)$ in $\mathbb{R}$ is said to converge to $x ∈ \mathbb{R}$, or $x$ is said to be a limit of $(x_n)$, if for every $\epsilon> 0$, there exists a natural number $K ∈ \mathbb{N}$ such that for all $n ≥ K$, the terms $x_n$ satisfy $\lvert x_n − x\rvert < \epsilon$.

Check to see if each of the following statements is equivalent to the definition of limit. If it is, prove it. If it is not, give an example to justify your conclusion.

1) For every $\epsilon > 0$, there exists a natural number $K ∈ \mathbb{N}$ such that for all $n > K$, the terms $x_n$ satisfy $\lvert x_n − x\rvert < \epsilon$.

2) For every $\epsilon \geq 0$, there exists a natural number $K ∈ \mathbb{N}$ such that for all $n ≥ K$, the terms $x_n$ satisfy $\lvert x_n − x\rvert < \epsilon$.

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    $\begingroup$ for 1) $K_1=K-1$ will work and for 2) $\epsilon=0$ will be a problem! $\endgroup$ – Bhauryal Sep 27 '14 at 7:27
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    $\begingroup$ 2) This condition is equivalent to the condition $x_n=x$ for all but finitely many $n$, which is clearly stronger than convergence (unless you space is discrete). $\endgroup$ – Jonas Dahlbæk Sep 27 '14 at 8:46
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  1. It is right. You can take $K_1=K-1$. Okay.

  2. It may be not right.

It may be right for the contant number, for example, $\{x_n: x_n=1\}$. However it is wrong for the following sequence$\{x_n: x_n=\frac1n\}$.

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  • $\begingroup$ No, not $K_1=K+1$. $\endgroup$ – Did Sep 27 '14 at 10:14
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    $\begingroup$ @Did: Yes. Thank You. $\endgroup$ – Paul Sep 27 '14 at 12:12
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1) is correct since you can choose $K_1=K-1$ and 2) is not true for $\epsilon=0$ as if you consider the sequence $\{x_n\}^{\infty}_{n=1}=\{\frac{1}{n}\}^{\infty}_{n=1}$ then you can't find any $K\in \Bbb{N}$ st $|x_n|<0 \quad \forall n>K$

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