A community for students.
Here's the question you clicked on:
 0 viewing
viniterranova
 3 years ago
Usint the epsilon delta definition evaluate limit x>3 x^2=9, where epsilon = 0,005.
viniterranova
 3 years ago
Usint the epsilon delta definition evaluate limit x>3 x^2=9, where epsilon = 0,005.

This Question is Open

yakeyglee
 2 years ago
Best ResponseYou've already chosen the best response.0Your question doesn't really make sense. The epsilondelta definition of the limit for a limit as \(x \rightarrow x_0\) says that there exists a real number \(L\), such that for all \(\epsilon>0\), there exists a \(\delta > 0\) such that if \(xx_0 < \epsilon\) then \(f(x)L<\delta\). The epsilondelta definition of the limit is only effective if you consider all \(\epsilon\), not a particular value of \(\epsilon\).

viniterranova
 2 years ago
Best ResponseYou've already chosen the best response.0Maybe this is because of the languague usage. The big chanllenge is to translate from the portuguese to English.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.