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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 |$X_n-L$|<$\epsilon$.

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    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. – 2017-02-20

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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.

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    That makes sense. Because the absolute value would always be 1, right? But the regular sequence would go -1, 1, -1, 1....? – 2017-02-20
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    Yes, you have it. Well done. – 2017-02-20
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    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? – 2017-02-20
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    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. – 2017-02-20
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    Would it be $\epsilon$ > |$X_n-L$| $\ge$ ||$X_n| - |L$||? Then we can simply use the definition of convergence to say that the $\lim \limits_{n \to \infty}$ |$X_n$| = |L|? – 2017-02-20
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    Johnnie, you're correct. $||x|-|L||\le |x-L|<\epsilon$. – 2017-02-20
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    Okay, sweet. For part c, I have that since $\lim \limits_{n \to \infty}$ |$X_n$| = 0, we know that ||$X_n$| - |0|| < $\epsilon$. Then, we know that ||$X_n$| - 0| = ||$X_n$|| = |$X_n$| = |$X_n$ - 0|, so |$X_n$ - 0| < $\epsilon$. Then we know that the $\lim \limits_{n \to \infty}$ $X_n$ = 0. Sound correct? – 2017-02-20
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    Yes, you have it. – 2017-02-21