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I have a question regarding when can limit sign be safely put inside of another function (e.g. absolute value) without changing any meaning.

I vaguely remember it has something to do with triangle inequality. For example, since integral does not follow triangle inequality, then $\int \lim f_n \ne \lim \int f_n.$

But I was wondering if $\lim |a_n| = |\lim a_n| $ in general (for $a_n$ are real numbers or functions)? And if this statement holds, how to prove it? If it does not hold, when can we move limit sign inside?

Alessio K
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yi G
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1 Answers1

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Assuming that $\lim_n a_n$ exists, then yes your formula is true. However it is possible that $\lim_n |a_n|$ exists but not $\lim_n a_n$, for instance if $a_n=(-1)^n$.

Let $$L=\lim_n a_n$$ By the triangle inequality $$\bigg| |a_n| - |L| \bigg|\leq |a_n-L|\rightarrow 0$$ hence $\lim |a_n|=|L|=|\lim_n a_n|$.

More generally if $f$ is a continuous function then $$\lim_n f(a_n)=f(\lim_n a_n)$$ In your case $f(x)=|x|$.

EDIT. Assume $f$ is continuous and that $L=\lim_n a_n$ exists. We actually only use that $f$ is continuous at $L$. Then if $\varepsilon>0$ is fixed there exists $\delta>0$ such that $$|x-L|<\delta\Rightarrow |f(x)-f(L)|<\varepsilon$$ Now since $a_n\rightarrow L$ we now that $|a_n-L|<\delta$ holds for large enough $n$, say $n>n_0$. Then if $n>n_0$ we have $$|a_n-L|<\delta\Rightarrow |f(a_n)-f(L)|<\varepsilon$$ $\varepsilon>0$ was arbitrary, hence $$\lim_n f(a_n)=f(L)$$ as claimed.

Olivier Moschetta
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  • As @player3236 noted in the comment above, the general equality only holds if both limits exist. It is possible for the left hand side to exist while the right hand side does not. – kccu Sep 22 '20 at 19:53
  • I suggest using \bigg| on the external absolute values in second formula. Could use \ell too for the limit. – zwim Sep 22 '20 at 19:55
  • Thank you so much! I understand how limit works with the absolute values now. But can you explain a little bit about why can one move limit inside when f is continuous? – yi G Sep 22 '20 at 19:57
  • Ok I'll edit. @zwim thanks I never know what commands are available in here. – Olivier Moschetta Sep 22 '20 at 19:58