Prove from first principles that if $x_n \rightarrow x$ as $n \rightarrow \infty$, then $|x_n| \rightarrow |x|$. I'm having trouble proving the limit of the above statement using first principles. Here is what I got so far, any tips you might have on continuing from here, would be much appreciated!
We are given $x_n \rightarrow x$ as $n \rightarrow \infty$.
$$n \geq N \rightarrow \big||x_n|-|x|\big| < \epsilon $$
Proving $ \big||x_n|-|x|\big| \leq|x_n - x|$.
$$ \big||x_n|-|x|\big| \leq |x_n-x| \iff\big||x_n|-|x|\big|^2 \leq|x_n-x|^2$$
$$\iff x_n^2 - 2|x_nx|+x^2 \leq  x_n^2 - 2x_nx + x^2 \iff x_nx \leq |x_nx| $$
Which is always true. Therefore...
$$\big||x_n|-|x|\big| \leq |x_n - x| < \epsilon$$
I'm not sure what to do after this as I have not practised much with a generalised sequences.
 A: After this, you take $\delta=\varepsilon$, and then$$\lvert x_n-x\rvert<\delta\iff\lvert x_n-x\rvert<\varepsilon\implies\bigl\lvert\lvert x_n\rvert-\lvert x\rvert\bigr\rvert<\varepsilon,$$since$$\bigl\lvert\lvert x_n\rvert-\lvert x\rvert\bigr\rvert\leqslant\lvert x_n-x\rvert.$$
A: That's fine, but this inequality only holds for real numbers. For complex numbers, use the triangle inequality $|a+b| \leq |a|+|b|$. Then, $|a| = |a-b+b| \leq |a-b|+|b| \Leftrightarrow |a|-|b| \leq |a-b|$. Similarly, $|b|-|a| \leq |b-a| = |a-b|$ holds and thus it follows that $||a|-|b|| = \max{|a|-|b|,|b|-|a|} \leq |a-b|$.
A: Let $ \varepsilon >0 $ :
Given that $ x_{n}\underset{n\to +\infty}{\longrightarrow} x\in\mathbb{R} $, there exists some $ n_{0}\in\mathbb{N} $, such that $ \left(\forall n\geq n_{0}\right),\ \left|x_{n}-x\right|<\varepsilon \cdot $
Since you've already proven that $ \left|\left|x_{n}\right|-\left|x\right|\right|<\left|x_{n}-x\right| $, we can use it to end with : $$ \left(\forall n\geq n_{0}\right),\ \left|\left|x_{n}\right|-\left|x\right|\right|<\varepsilon $$
Therefore $$ \left|x_{n}\right|\underset{n\to +\infty}{\longrightarrow}\left|x\right| $$
