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Given $f: \mathbb{R} \to \mathbb{R}$

Is finding the limit of the function $\lim_{x \to \infty} f(x)$ equivalent to find the limit of the sequence $\{f(n)\}_{n = 1}^\infty$? Intuitively, this seems obvious, but Rudin defines limits as $x$ approaches points in the extended real number system as follows

$$f(t) \to A \\\ \text{as} \\\ t \to x$$

where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ , there exists a neighborhood $V$ of $x$ such that $V \cap E$ is not empty, and such that $f(t) \in U$ for all $t \in V \cap E, t \ne x$

If infinite limits can be regarded as limit of the sequence $f(n)$, $n = 1, 2 ...$, how can I prove that this definition coincides with the definition of the convergence of a sequence (i.e for every $\epsilon > 0$, there exists $N$....)?

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They are not equivalent. Consider the function: $$ f(x) = \sin(\pi x). $$

The limit of the function does not exist while the limit of the sequence f(n) is zero.

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  • $\begingroup$ Thanks, I wasn't thinking clearly. $\endgroup$ – user59298 Jan 23 '13 at 23:32

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