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?
Exchanging Limits of Functions/Limits of Sequences
2 Answers
A classical epsilon-delta proof, if ever there was one... Here we go:
Let $\varepsilon\gt0$. Since $f$ has limit $L$ at $p$, there exists $\delta\gt0$ such that, for every $x$, $|x-p|\leqslant\delta$ implies $|f(x)-L|\leqslant\varepsilon$.
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 $n\geqslant N$.
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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0The 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?). – 2012-03-13
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.