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Let $(x_n) \subset \mathbb{R}$ be a sequence of real numbers and $l \in \mathbb{R}$
Define $f:\mathbb{N} \rightarrow \mathbb{R}, n \mapsto f(n):=x_n$
Prove that $\displaystyle\lim_{y\to\infty}f(y)=l$ iff $\displaystyle\lim_{n\to\infty}x_n=l$

I have no idea where to begin; I'm not even sure why we can discuss $\displaystyle\lim_{y\to\infty}f(y)$

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  • $\begingroup$ We can discuss $lim_{x\rightarrow +\infty} f(x)$ by taking larger and larger values of $x$ and seeing if $f(x)$ approaching a fixed value $c$ (in this case, $lim_{x\rightarrow +\infty} f(x)=c$ ), if it goes to infinite, that is, $lim_{x\rightarrow +\infty} f(x)=\pm \infty$, or if it doesn't have a limit after all (like $lim_{x\rightarrow +\infty} sin(x)$. $\endgroup$ – Marra Feb 24 '13 at 20:39
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    $\begingroup$ $f(x_n)$ dont have sense, the $f$ domain is natural numbers $\endgroup$ – Gaston Burrull Feb 24 '13 at 20:40
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    $\begingroup$ Ok, problem is just a change of name, replace $f(n)$ by $x_n$ (by definition of $f$), there is nothing to prove $\endgroup$ – Gaston Burrull Feb 24 '13 at 20:44
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    $\begingroup$ What you have so far is something like set $x=y$ prove that $x=1$ iff $y=1$ $\endgroup$ – leo Feb 24 '13 at 20:46
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    $\begingroup$ maybe it could help to take a look at the definition of a sequence which is exactly a mapping say $g:\mathbb{N}\to X$ for some space $X$ with $i\mapsto g(i)=:g_i$. Hence I would agree that there is nothing to proof since your mapping $f$ is defined by the sequence $\endgroup$ – Quickbeam2k1 Feb 24 '13 at 20:57
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First we have,

$$\lim_{y\to \infty} f(y)=\lim_{n\to \infty} f(n)$$

since the name of variable does not change the limit definition (as you can check with $\epsilon$-$\delta$ usual definition). By definition of the function $f$ we have that $f(n)=x_n$ for all $n\in\mathbb{N}$ then $$\lim_{y\to \infty} f(y)=\lim_{n\to \infty} f(n)=\lim_{n\to \infty} x_n$$

finally $$\lim_{y\to \infty} f(y)=l\Leftrightarrow \lim_{n\to \infty} x_n=l.$$

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