# Prove validation of if $\lim_{n\to\infty}\{x_n\} = a$ then $\lim_{n\to\infty}\{|x_n|\} = |a|$

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?

• 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, $|x| \to |a|.$" – Will M. Nov 23 '18 at 16:31
• @WillM. indeed, you hint makes it easier. Thanks for taking your time! – roman Nov 23 '18 at 16:37