How is $\lim_{x\to a}|f(x)| = |\lim_{x\to a}f(x)|$?

Question

The title basically says it all. My book states that $$ \lim_{x\to a}|f(x)| = |\lim_{x\to a}f(x)|\ $$ in proving that, if $f(x)$ is continuous at $a$, so is $|f(x)|$. Why is this true?

Answer 1

This is basically a direct application of two facts:

1. The absolute value function $|\cdot|:\mathbb{R}\to[0,\infty)$ is continuous;
2. The composition of continuous functions is continuous.

The second fact is very standard. The first is a consequence of the reverse triangle inequality $\big||x|-|y|\big|\leq|x-y|$: for $\varepsilon>0$, if $\delta=\varepsilon$, then for $|x-x_0|<\delta$: $$\big||x|-|x_0|\big|\leq|x-x_0|<\varepsilon,$$ so $|\cdot|$ is continuous at $x_0$ for any $x_0\in\mathbb{R}$.

From this, we can conclude that if $f$ is continuous at $x_0$, then: $$\lim_{x\to x_0}|f(x)|=\left|\lim_{x\to x_0}f(x)\right|,$$ by the fact that continuity implies sequential continuity; that is, if $x\to x_0$ and $g$ is continuous at $x_0$, then $g(x)\to g(x_0)$. (Here we're applying the fact that $|\cdot|$ is continuous at $f(x_0)$.)