TITLE: Limits and absolute values
QUESTION [0 upvotes]: Let $f:E \to \mathbb{R}$. Let $a \in E$ such that $\lim_{x \to a} f(x)$ exists. Show that $\lim_{x\to a} |f(x)|$ exists and the following identity holds:
$$
\lim_{x\to a} |f(x)| = \left| \lim_{x\to a} f(x)\right|.
$$

REPLY [0 votes]: Since the absolute value is a continuous function, the limit commutes, i.e. the absolute value of the limit will be the limit of the absolute value, provided it exists. In general, limits commute with continuous functions; that is to say if $S$ and $T$ are sets of numbers, and $f:S\to T$, $g:T\to \mathbb{R}$ are functions, with $a\in S$ and $b=f(a)$, such that $f$ is continuous at $a$ and $g$ is continuous at $b$, then $g\circ f$ is continuous at $a$. In other words,
$$ \lim_{x\to a}g(f(x)) = g(\lim_{x\to a}f(x))$$
To prove this, let $\epsilon>0$, there exists $\delta>0$ such that if $y\in T$ and $|y-b|<\delta\implies |g(y)-g(b)|<\epsilon$ since $g $ is continuous. For the $\delta$ just found, there exists $\delta_1$ such that if $x\in S$ and $$|x-a|<\delta_1\implies |f(x)-b|<\epsilon$$ So $|x-a|<\delta_1\implies |g(f(x))-g(f(a))|<\epsilon$
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