# Real Analysis: Proof of an absolute value of a function exists

Suppose $\lim_{x\to a} f(x) = L$. Prove that $\lim_{x\to a} \lvert f(x)\rvert$ exists.

The $\epsilon-\delta$ definition of limits may be helpful. My logic just cannot get there without feeling incomplete. I tried proof by contradiciton and including the squeeze theorem. I really just can't get a grasp for how to think of proofs. Any help is appreciated, thanks!

If we can prove that $||f(x)| - |L|| \leq |f(x) - L|$ for all $x$, and if we use the assumption, then for every $\varepsilon > 0$ there is some $\delta > 0$ such that $0 < |x-a| < \delta$ implies $||f(x)| - |L|| \leq |f(x)-L| < \varepsilon$; then $|f(x)| \to |L|$ as $x \to a$.
The target inequality is actually not deep, which follows directly from triangle inequality. Let $a,b$ be real numbers. If we use the triangle inequality, then $|a| = |a-b+b| \leq |a-b| + |b|$ and $|b| = |b-a+a| \leq |a-b| + |a|$. So $|a| - |b| \leq |a-b|$ and $|b| - |a| = -(|a| - |b|) \leq |a-b|$; so $||a| - |b|| \leq |a-b|$.
$\big||f(x)|-|L|\big|\leq|f(x)-L|$, can you finish from here?
• $|a+b|\leq|a|+|b|$ and realise $a=f(x)-L$ and $b=L$, one gets $|f(x)|-|L|\leq|f(x)-L|$ and realise $a=L-f(x)$ and $b=f(x)$ to get $|L|-|f(x)|\leq|f(x)-L|$. – user284331 Feb 23 '18 at 3:23