Prove that $\lim(x_n)=0$ if and only if $\lim(|x_n|)=0$.

Definition: Let $X = (x_n)$ be a sequence in $\mathbb{R}$ and let $x\in\mathbb{R}$.

Then $\lim(x_n) =x$ iff for all $\varepsilon>0$, $\exists k\in\mathbb{N}$ such that $|x_n-x|<\varepsilon$ for all $n\geq k$.

I am wondering if this is sufficient:

If we know that $\lim(x_n)=0$, we know for all $\varepsilon>0$, $\exists k\in\mathbb{N}$ such that $|x_n-0|<\varepsilon$ for all $n\geq k$.

We can rewrite this as the following: $$|x_n-0|<\varepsilon \Leftrightarrow ||x_n|-0|<\varepsilon\Leftrightarrow \lim(|x_n|)=0$$

Then the conclusion seems to follow logically. Is the proof really that simple or are there some things I am missing?

  • 2
    $\begingroup$ That's exactly right. $\endgroup$ – Cameron Williams Oct 2 '15 at 1:25
  • $\begingroup$ Your solution seems fine to me,Bravo! $\endgroup$ – Arpit Kansal Oct 2 '15 at 2:20

Your proof is correct. Note that we also have (in general)

$$\lim x_n = c \Rightarrow \lim |x_n| = |c|$$

and that $\lim |x_n| = c$ does not in general imply that $\lim x_n$ exists.


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