Finding the limit of a sequence vs function. 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 $N$....)?

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.
