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How would I do this using the definition of continuity for all real values of $|x|$?

I know there is a proof that you can use that a function is continuous if and only if the limit at that point is equal to $f(a))$ for example, but that would only tell me that it is continuous at $x=0$ for example.

If proving at $x=0$, I would say that for all $\epsilon$ > $0$, there exists a $\delta$ > $0$, such that when $|x|$ < $\delta$, then $| |x| | = |x| < \delta$ < $\epsilon$, when $\delta$ = $\epsilon$.

Now for all real numbers $a$: for all $\epsilon$ > $0$, there exists a $\delta$ > $0$, such that when $|x-a|$ < $\delta$, $| |x-a| | <= |x-a|$ using the reverse triangle inequality < $\delta$ < $\epsilon$, when $\delta$ = $\epsilon$.

Edit: I made this attempt, not sure how legit it is.

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    Sorry, I posted my answer without seeing your edits. Your solution with the $\varepsilon - \delta$-definition is correct. But it would also work with the equivalent characterisation of continuity with limits. – Lukas Oct 11 '22 at 19:10
  • @Lukas Cheers mate. Reading my answer I don’t know why I couldn’t do it at first, it’s actually not that hard. Now I have to work out how to prove continuity of harder functions like sign functions etc :/ –  Oct 11 '22 at 19:11

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We can use the usual $\varepsilon-\delta$-definition.

Let $\varepsilon >0$. Choose $\delta = \varepsilon$. Then we have for any $x_0 \in \mathbb R$ that $$||x|-|x_0|| \le|x-x_0| < \delta = \varepsilon$$ for all $x$ such that $|x-x_0|<\delta$. Therefore $f(x) = |x|$ is continuous in any $x_0 \in \mathbb R$. (Where we used the "reverse" triangle inequality for subtractions: $|x-y| \ge ||x|-|y||$ for any $x,y \in \mathbb R$ which can be derived from the usual one)

You could also use the equivalent characterisation that you mentioned with the limits: For $x_0=0$ we have $$\lim_{x \to 0-} |x| = \lim_{x \to 0-} -x =0 =f(0) =\lim_{x \to 0+}x = \lim_{x \to 0+} |x|$$ and for $x_0>0$ we have that $|x| = x$ for all $x$ sufficiently near to $x_0$ which also gives $$\lim_{x \to x_0-} |x| = \lim_{x \to x_0-} x =x_0 =f(x_0) =\lim_{x \to x_0+}x = \lim_{x \to x_0+} |x|$$ and similarly for $x_0<0$ with $|x| = -x$ for all $x$ sufficiently close to $x_0$. This way we see that at every point the right and left limits are the same, so the limits exist and they coincide with the value $f(x_0)$. This is another way of showing $f$ is continuous.

Lukas
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