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Can we always interchange limits and the absolute value function? That is, is $$\lim_{n \to \infty} |x_n|$$ always equal to $$|\lim_{n \to \infty} x_n|$$ where $x_n$ is a sequence of real numbers?

ManUtdBloke
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    Are you assuming that both exist? Otherwise, let $x_{2n}=1$ and $x_{2n+1}=-1$. Then the top limit is $1$ but the bottom limit fails to exist. – lulu Sep 20 '16 at 10:43
  • I just noticed the author of the answer to this post - http://math.stackexchange.com/questions/71121/space-of-bounded-continuous-functions-is-complete - used it a few times. How is it justified in his case? – ManUtdBloke Sep 20 '16 at 10:46
  • Have you written to the writer of that post to ask for details? A lot of these problems go away if you know that all the limits in the picture exist. Maybe existence is guaranteed in the context you cite. – lulu Sep 20 '16 at 10:49
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    This is true if $\lim_{n\rightarrow\infty}x_n$ exists, since $|\cdot|$ is a continuous function. – Marc Sep 20 '16 at 10:53

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If $x_n$ is convergent, then it always holds, i.e.,
$$\lim_{n \to \infty} |x_n|=|\lim_{n \to \infty} x_n|.$$

This bacause that $f(x)=|\cdot|$ is a continuous function, and hence we have $$\lim_{n\to\infty}f(x_n)=f(\lim_{n\to \infty}x_n).$$

Paul
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  • So it holds due to either continuity of the function $f(x) = |\cdot|$ or convergence of the sequence $x_n$? Or are both properties required? I'm trying to understand precisely why it holds? – ManUtdBloke Sep 20 '16 at 10:56
  • @eurocoder It only needs the convergence of $x_n$. $|\cdot|$ is continuous (This is a fact). – Paul Sep 20 '16 at 10:59
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No; the easiest example is to consider $x_n=(-1)^n$. We have $$\lim_{n\to\infty}|x_n|=1$$ while the sequence $x_n$ diverges as $n\to\infty$.

Clayton
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  • Ok thanks. I noticed the author of the answer to this post - http://math.stackexchange.com/questions/71121/space-of-bounded-continuous-functions-is-complete - used it a few times. How is it justified in his case? – ManUtdBloke Sep 20 '16 at 10:47
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    @eurocoder Because $f$ is by definition the limit of $f_n$ which is a convergent sequence, i.e. $\lim_{n\to \infty} f_n = f.$. – Eff Sep 20 '16 at 11:07
  • @Elf The author of the answer to that post doesn't use that property when stating: $|\lim_{n \to \infty} f_n(x) - f_N(x)| = \lim_{n \to \infty} |f_n(x) - f_N(x)|$. So how does he justify it in this case? – ManUtdBloke Sep 20 '16 at 12:56