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.
 A: I propose another approach, which is more complicated but also more general: to prove the continuity of $|f(x)|$, note that $|f(x)|=g(f(x))$ where $g(x)=|x|$ and recall that if $f,g$ are two functions such that $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g(f(x))$ is continuous at $a$. The proof of this assertion is routine $\epsilon$-$\delta$.
Proof of the previous assertion: for every $\epsilon>0$, since $g$ is continuous at $f(a)$ there is a $\delta_1 >0$ such that if $|f(x)-f(a)| < \delta_1$, then 
$$ |g(f(x))-g(f(a))| \le \epsilon. $$
Since $f$ is continuous at $a$, there exists a $\delta_2 > 0$ such that if $|x-a| < \delta_2$, then
$$ |f(x)-f(a)| \le \delta_1. $$
In the last part I choose $\epsilon= \delta_1$ in the definition of continuity of $f$ at $a$, this is the key passage that provides the connection between the two limit definitions we are given. Since $\epsilon$ is arbitrary, note that this is exactly the definition of continuity of $g(f(x))$ at $a$.
A: First, show that the function $g(x)=|x|$ is continuous on reals, then use the fact that the composition of two continuous functions is also continuous ($|f(x)|= g \circ f (x)$).
A: 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?
