Prove that if $\lim_{n\to \infty} a_{n} = L$. Then $\lim_{n\to \infty}|a_{n}| = |L|$ If $\lim_{n\to \infty} a_{n} = L$.
Then $\lim_{n\to \infty}|a_{n}| = |L|$ i.e. $\forall$ $\epsilon>0$, $\exists n_{0} \in \mathbb{N}$, $\forall n\geq n_{0}$ $| |a_{n}|-|L||<\epsilon$.
Because of $\lim_{n\to \infty} a_{n} = L$. I know that $\forall$ $\epsilon>0$, $\exists n_{0} \in \mathbb{N}$, $\forall n\geq n_{0}$ $|a_{n}-L|<\epsilon$. This looks quite similar to what I need to prove. 
So I thought that, by giving a certain $\epsilon>0$ (maybe $|L|>0$ or $\epsilon + |L|>0$) and using that $|a_{n}|-|L|\leq|a_{n}-L|<\epsilon$. I could get to the solution. However, I can't really seem to get past this point in order to get to the expression that would complete my proof (that being  $| |a_{n}|-|L||<\epsilon$.)
Any help would be very much appreciated.
 A: Try this proof.
Let $\{a_n\}$ be a convergent sequence to some limit $L$. We prove that $\{|a_n|\} \to |L|$. Given $\epsilon > 0$, choose $N$ so that for any $n \geq N$, $|a_n - L| < \epsilon$. (We may do this because $\{a_n\} \to L$.) For any $n \geq N$, we have: 
$$||a_n| - |L|| \leq |a_n - L| < \epsilon.$$
Hence, $\{|a_n|\}$ converges to $|L|$.
Note that $||x| - |y|| \leq |x-y|$ is a fairly easy lemma to prove that follows directly from the normal triangle inequality. Upon request, here is a proof. 
Lemma. For $x, y \in \mathbb{R}$, $||x| - |y|| \leq |x-y|$.
It suffices to prove that 
$$-|x-y| \leq |x| - |y| \leq |x-y|.$$
For the second inequality, we have: 
$$|x| = |x + 0| = |(x-y) + y| \leq |x-y| + |y|.$$
Rearranging gives $|x| - |y| \leq |x-y|$. 
Similarly, we have: 
$$|y| = |y + 0| = |(y - x) + x| \leq |y-x| + |x| = |x-y| + |x|.$$
Rearranging gives $|y| - |x| \leq |x-y|$. Multiplying through by $-1$ and reversing the inequality sign gives the desired 
$$-|x - y| \leq |x| - |y|.$$
A: Function $\ A:\Bbb R\to\Bbb R\ $ given by:
$$ \forall_{x\in\Bbb R}\quad A(x)\ :=\ |x| $$
is Lipschitz with constant $1$ (it doesn't stretch any distance) hence function $\ A\ $ is continuous, which means that $\ \lim_{n\to\infty}\ |a_n|\,=\,|a|\ $ for every sequence $\ (a_n)\ $ and real $\ a\ $ such that
$\ \lim_{n\to\infty}\ a_n\,=\,a.\,\ $ Great!
