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

Question

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?

Answer 1

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.