How to show that $f(x)=|x|$ is continuous? Show that the function $f:\mathbb{R}\to\mathbb{R}$, defined by $f(x)=|x|$ is continuous. I only know how to prove the function is continuous at some point. How to prove this one? To prove that it is continuous at all $x_0$?
 A: If you want to use the $\varepsilon$-$\delta$ definition of continuity, then the inequality $||x|-|y||\leq |x-y|$ is the best place to start.
A: That a function is continuous simply means that it is continuous at every point of its domain.
A trivial $\epsilon,\delta$-proof goes as follows.
Let $x_0 \in \mathbb{R}$ be arbitrary. Let $\epsilon > 0$. Let $\delta = \epsilon > 0$. Then for any $x \in \mathbb{R}$ with $0 < |x-x_0| < \delta$, we trivially have $||x|-|x_0|| \leq |x-x_0| < \epsilon$ by our choice of $\delta = \epsilon$. 
The inequality $||x|-|x_0||\leq|x-x_0|$ is simply the reverse triangle inequality.
A: For all the $x_0 > 0$  you have in a neighbourhood of $x_0$ that $|x|=x$ and you know $f(x)=x$ is continuous over $\mathbb{R}$. The same goes for $x_0<0$. Then you only have to prove the continuity in $0$. Considering that:
-$\lim_{x\rightarrow 0^-} |x|=\lim_{x\rightarrow 0^-} -x=0$
-$\lim_{x\rightarrow 0^+} |x|=\lim_{x\rightarrow 0^+} x=0$
you can conclude that $$\lim_{x\rightarrow 0} |x|=0 $$
therefore the function is continuous in $0$.
EDIT to prove that a function is continuous you just have to prove that it is continuous in every point (in fact, that's the definition)
