Show that $|f(x)|$ is continuous at $a$ 
Suppose that $f(x)$ is continuous at $a$. Show that the function $|f(x)|$ is continuous at $a$.

Proof:
Since $f(x)$ is continuous at $a$ then, $$\lim_{x\to a}f(x)=f(a)$$
Show that $|f(x)|$ is continouos at $a$ $$\lim_{x\to a}|f(x)|=...$$
From here I can not figure a way to finish the proof. In my head $|f(x)|$ might not be continuous a $a$, such as if $f(a)$ is negative. Then $|f(a)|$ would be positive. Any help would be appreciated! Preferably relating to the Basic Limit Theorems of continuity if possible.

I claim it converges to $|f(a)|$. Taking any sequence $x_n \to a$, we get by the inverse triangle inequality that 
$$
0 \leq| \; |f(x_n)|-|f(a)| \; | \leq |f(x_n)-f(a)|
$$
Now the right-hand side goes to 0, therefore we can use the sandwich lemma to conclude that $|f(x_n)| \to |f(a)|$  
Edit: How does this proof continuity? 
We now know that $\lim_{n \to \infty}|\; |f(x_n)|-|f(a)|\;|=0$
For limits in the real numbers, we know that $\lim_{n \to \infty}|x_n -x|=0$ $\iff$ $\lim_{n \to \infty}x_n=x$. If you have not proven this in your class, its a good and simple exercise to proof it yourself.
Edit: Also a $\delta - \epsilon $ proof. Let $\epsilon > 0$ be given. Then there exists a $\delta > 0$ s.t. $|x-a|< \delta$ => $|f(x)-f(a)| < \epsilon$. Can you finish the proof using the inverse triangle inqueality again?
