# Exchanging Limits of Functions/Limits of Sequences

For a function $f$, if the limit $$\lim_{x \to p}f(x) = L$$ exists and $(x_n)$ is a sequence that converges to $p$, I'm trying to figure out whether $$\lim_{x \to p}f(x) = \lim_{n \to \infty}f(x_n)$$ and prove if true or give a counterexample. I believe it is true because since the (first) limit exists, I can pick $\epsilon > 0$ such that $$|x - p| < \delta \implies |f(x) - L| < \epsilon$$ and since the sequence converges to $p$, I can find $N$ such that $$n > N \implies |x_n - p| < \epsilon$$ If I set $\delta = \epsilon$, the first implication can be rewritten such that for all $n > N$, $$|x_n - p| < \delta \implies |f(x_n) - L| < \epsilon$$ which means that $$\lim_{n \to \infty}f(x_n) = L.$$ Does this approach look OK?

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A classical epsilon-delta proof, if ever there was one... Here we go:

1. Let $\varepsilon\gt0$. Since $f$ has limit $L$ at $p$, there exists $\delta\gt0$ such that, for every $x$, $|x-p|\leqslant\delta$ implies $|f(x)-L|\leqslant\varepsilon$.

2. The sequence $(x_n)$ converges to $p$ hence there exists a finite $N$ such that $|x_n-p|\leqslant\delta$ for every $n\geqslant N$. In particular, $|f(x_n)-L|\leqslant\varepsilon$ for every $n\geqslant N$.

3. Since $\varepsilon$ is as small as desired, this proves that $(f(x_n))$ converges to $L$, since one showed: $$\forall\varepsilon\gt0,\quad\exists N,\quad\forall n\geqslant N,\quad |f(x_n)-L|\leqslant\varepsilon.$$

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There is no assumption that $f$ is continuous at p... – AFX Mar 13 '12 at 21:31
@AFX True. Modified. Thanks. – Did Mar 13 '12 at 21:33
So, how is this proof different than the one I gave? Maybe I'm overlooking something but the logic appears basically the same. – AFX Mar 13 '12 at 21:35
The logic is roughly the same, yes, but there are some odd steps in your post: pick epsilon?? (one does not pick epsilon, one tries to show something for EVERY epsilon) // delta appearing from nowhere // set delta=epsilon (sorry?). – Did Mar 13 '12 at 22:28

Yes.

The only thing I would mention is to be careful with your "$\delta=\epsilon$", as the $\epsilon$ in that sentence is not the $\epsilon$ you are carrying in your proof.

Also, at the beginning, you are not picking $\epsilon$. Given an $\epsilon>0$, you can pick a $\delta$ such that the inequality holds.

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