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The sequence $x_n=2+1/n$ certainly gets "closer and closer" to $2$ as $n$ gets "larger and larger." And we know that $$\lim_{n\to\infty}2+1/n=2$$ What's going on here? Use the definition to explain what we really mean when we say a sequence gets "closer and closer" to the limit.

I'm not sure how to explain it with the definition..

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Why not tell us what definition of limit you are using? – Henry May 12 '12 at 23:55

The definition, put into words, says that you can guarantee that the term $2+(1/n)$ of the sequence will be as close to 2 as you like, simply by taking $n$ large enough.

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This is effectively the $\varepsilon$-$\delta$ notion of the limit... – J. M. May 14 '12 at 2:55
    
Yes (or maybe the $\epsilon-N$ notion); it isn't entirely clear to me just what OP wants to know, but I thought it might have something to do with expanding the usual definition into simple declarative English. – Gerry Myerson May 14 '12 at 3:14

That $2+\frac{1}{n}\to 2$ means that $$\forall\epsilon>0\,\,\exists N_\epsilon\in\mathbb{N}\,\,s.t.\,\,n>N_\epsilon\Longrightarrow \left|2+\frac{1}{n}-2\right|=\left|\frac{1}{n}\right|<\epsilon$$

The last inequality actually means $\,\,-\epsilon<\frac{1}{n}<\epsilon\,\,$ for all the indexes n greater than that "special" index $\,\,N_\epsilon\,\,$, which we can convey colloquially as that the original sequence "gets closer and closer" (perhaps it'd be better to say "it gets arbitrarily closer") to its limit.

DonAntonio

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user31254, you might like to choose a user name, rather than signing your answers each time. – Antonio Vargas May 13 '12 at 23:09
    
Yes, but I just don't know how to do it...:) – DonAntonio May 14 '12 at 2:01

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