Given a convergent sequence $\{x_n\}$: $$ \lim_{n\to\infty}\{x_n\} = a $$ Prove that $\{|x_n|\}$ is also convergent and: $$ \lim_{n\to\infty}\{|x_n|\} = |a| $$

I've started with the definition. If limit of $x_n$ exists then for some $\varepsilon > 0$ we have that: $$ |x_n - a| < \varepsilon $$

Now consider the sequence involving absolute values. If it does converge then: $$ ||x_n| - |a|| < \varepsilon $$

But by triangular inequality we have that: $$ |a+b| \le |a| + |b| \iff |a+b| - |b| \le |a| $$ Let $c = a + b$ hence $a = c - b$: $$ |c-b+b| - |b| = |c| - |b| \le |c-b| \iff \\ \iff||c| - |b|| = \pm(|c| - |b|) \le |c-b| $$

So we have that: $$ ||x_n| - |a|| \le |x_n - a| <\varepsilon $$

Thus going back from $\varepsilon$ definition to a limit we may conclude that: $$ \lim_{n\to\infty}\{x_n\} = a \implies \lim_{n\to\infty}\{|x_n|\} = |a| $$

I'm not sure whether the above is a correct reasoning, so could someone please verify this?

  • $\begingroup$ The proof is correct. You may want to rephrase it as: "Since $|x| \leq |x-a|+|a|$ and $|a| \leq |x-a|+|x|$ we see at once $||x|-|a||\leq|x-a|$ hence if $|x-a|\leq\varepsilon$ we get $||x|-|a||\leq \varepsilon,$ that is, $|x| \to |a|.$" $\endgroup$ – Will M. Nov 23 '18 at 16:31
  • $\begingroup$ @WillM. indeed, you hint makes it easier. Thanks for taking your time! $\endgroup$ – roman Nov 23 '18 at 16:37

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