Prove that if $\lim \limits_{n \to \infty}$ $x_n$ = $L$, then $\lim \limits_{n \to \infty}$ $|x_n|$ = $|L|$.

This question has two other parts.

b) Give an example of a sequence $(x_n)$n∈$\mathbb{N}$ such that (|$x_n$ |)n∈$\mathbb{N}$ converges, but ($x_n$)n∈$\mathbb{N}$ diverges.

c) Prove that if $\lim \limits_{n \to \infty}$ $|x_n|$ = 0, then $\lim \limits_{n \to \infty}$ $x_n$ = 0.

I'm pretty sure we're supposed to use our definition for convergences, which says that Given a real number $L$, we say that $(X_n)$ converges to L if for every $\epsilon$>0, there exists N∈$\Bbb{N}$ such that for all n∈N satisfying n>N, we have |$X_n-L$|<$\epsilon$.

• Also, I just joined math stack exchange a few weeks ago and you all have been a big help. I'm slowly understanding the formatting, let me know if I did anything wrong. Feb 20, 2017 at 15:54

1 Answer

For the first part, just use the "left-sided" triangle inequality in the $\delta-\epsilon$ definition.

For part b), the classical example is the sequence $a_n=(-1)^n$.

Part c) is trivial given $||x_n||=|x_n|<\epsilon$ implies convergence.

• That makes sense. Because the absolute value would always be 1, right? But the regular sequence would go -1, 1, -1, 1....? Feb 20, 2017 at 16:03
• Yes, you have it. Well done. Feb 20, 2017 at 16:04
• Sorry, I didn't realize you added answers to the other questions. What do you mean use the left sided triangle inequality? We can assume that $\lim \limits_{n \to \infty}$ $x_n$ = $L$. From that, I know that |$X_n-L$|<$\epsilon$. Then what? Feb 20, 2017 at 19:49
• Okay, so we know |$X_n-L$|<$\epsilon$. By the triangle inequality, we know that $\epsilon$ > |$X_n-L$| $\ge$ |$X_n$| - |$L$|. Right? I don't see where to go from there. Feb 20, 2017 at 20:10
• Johnnie, you're correct. $||x|-|L||\le |x-L|<\epsilon$. Feb 20, 2017 at 20:38