1
$\begingroup$

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?

$\endgroup$
1
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.