# $\varepsilon$-$\delta$-proof that $3|x| − 2$ is continuous at $x = 0$

I am having a lot of issues figuring out how to use the formulas: $$|f(x) − f(y)| < \varepsilon$$ and $$|x − y| < \delta$$ when finding the proof that the function $$f(x) = 3|x| − 2$$ is continuous at $$x = 0$$. What are the steps to answering this type of question?

• JoseSquare's answer is very good, and if you want to see another example where they go through the steps for finding continuity, I would check here: math.umd.edu/~mboyle/courses/410f12/uniform.pdf Feb 15 '19 at 21:44

You have to proof that at $$x=0$$ for any given $$\epsilon > 0$$ there exist $$\delta >0$$ such when $$|x|<\delta \Rightarrow |f(x) - f(0)|< \epsilon$$, so you want that $$|3|x|-2 -(-2)|= |3|x||<\epsilon$$. So $$3|x| < 3\delta$$, so if you take $$3 \delta < \epsilon$$, which is $$\delta < \frac{\epsilon}{3}$$, then all the hypotesis of continuity at $$x=0$$ are veryfied. This is normally the way to proceed in this type of problems.