## anonymous 4 years ago Find a number δ > 0 such that for all x, 0 < absolute value (x-x0) < δ → absolute value (f(x) - L) < ϵ f(x)=4/x, x0=2, ϵ=0.4 Please explain in detail. Unfortunately, my school took a long time to fix my schedule so I was not put into AP BC Calculus until 2 or so weeks in today so much of the foundation is missing but I'm studying on my own to catch up.

1. anonymous

2. anonymous

they are just variable they use greek letters to annoy you

3. anonymous

|dw:1348198061023:dw|

4. anonymous

What do those variables represent?

5. anonymous

|dw:1348198204753:dw|

6. anonymous

δ Is delta right? But does it mean the same as the triangle delta?

7. anonymous

what is the triangle delta ?

8. anonymous

you want to make $$f$$ close to $$L$$ how close? within $$\epsilon$$ for arbitrary $$\epsilon$$ in your case (\epsilon\) is not arbitrary, it is 0.4

9. anonymous

delta can be 02 or less to satisfy .

10. anonymous

Δ I mean. Seems it's upper case delta.

11. anonymous

it is just a variable, you could use anything $$a$$, $$b$$, $$\alpha$$ $$\beta$$ etc

12. anonymous

Ok. I'm frantically flipping through my textbook trying to comprehend all this.

13. anonymous

In this case, what would epsilon be representing specifically?

14. anonymous

What is the name of these parts of calculus?

15. anonymous

Also, my textbook says δ = 1/3 is the answer but how they got it I have no idea.