## j814wong Group Title 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. one year ago one year ago

1. j814wong Group Title

Also, please explain the symbols.

2. satellite73 Group Title

they are just variable they use greek letters to annoy you

3. mahmit2012 Group Title

|dw:1348198061023:dw|

4. j814wong Group Title

What do those variables represent?

5. mahmit2012 Group Title

|dw:1348198204753:dw|

6. j814wong Group Title

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

7. mahmit2012 Group Title

what is the triangle delta ?

8. satellite73 Group Title

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. mahmit2012 Group Title

delta can be 02 or less to satisfy .

10. j814wong Group Title

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

11. satellite73 Group Title

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

12. j814wong Group Title

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

13. j814wong Group Title

In this case, what would epsilon be representing specifically?

14. j814wong Group Title

What is the name of these parts of calculus?

15. j814wong Group Title

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