In a limit proof, what are the assumptions? In a proof.
Prove that given:
$$\lim_{x \to a} f(x) = L$$ then
$$\lim_{x\to a} |f(x)| = |L|$$
We know that
$$|f(x) - L| < \epsilon \space \text{for} \space |x - a| < \delta_1$$
What is the objective then?
Do we prove there exists a $\delta_2$ such that $\displaystyle \lim_{x\to a} |f(x)| = |L|$
Or do we go from the fact thatit is true that: $|f(x) - L| < \epsilon \space \text{for} \space |x - a| < \delta_1$ and then somehow using $\delta_1$ derive that:
$| |f(x)| - |L| | < \epsilon$
Or do we find a $\delta$?
Thanks!
 A: You need to show that $\forall \epsilon >0, \ \exists \delta_2 > 0$, such that
$$
| |f(x)| - |L|| \ < \epsilon \quad \forall |x-a| < \delta_2
$$
A: You always need to define the $\delta$ with respect to a given $\epsilon$, and you need to 
relate to $\delta_1$ as well.
In this case, the relationship should be easy once you notice the following fact:
$$||a| - |b|| \leq |a - b| \quad \forall a,b\in \mathbb{R}$$
A: By assumption the function $g$ obtained from $f$ by putting $g(x):=f(x)$ when $x\ne a$ and  $g(a):=L$ is continuous at $a$. Therefore the function $|g|:={\rm abs}\circ g$ is continuous at $a$ as well.
A: Since you are given $$\lim_{x \to a} f(x) = L$$ then you already know anytime $|x-a|<\delta_1$ that $|f(x)-L|<\varepsilon$. By the reverse triangle inequality you also have $$\left||f(x)|-|L|\right|\leq |f(x)-L|$$ which means anytime $$|x-a|<\delta_1 \quad \text{then} \quad \left||f(x)|-|L|\right|<\varepsilon \\ \implies \lim_{x \to a}|f(x)|=|L|$$ so you don't need to find a $\delta_2$, as $\delta_1$ suffices.
