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j814wong
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.
Also, please explain the symbols.
they are just variable they use greek letters to annoy you
|dw:1348198061023:dw|
What do those variables represent?
|dw:1348198204753:dw|
δ Is delta right? But does it mean the same as the triangle delta?
what is the triangle delta ?
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
delta can be 02 or less to satisfy .
Δ I mean. Seems it's upper case delta.
it is just a variable, you could use anything \(a\), \(b\), \(\alpha\) \(\beta\) etc
Ok. I'm frantically flipping through my textbook trying to comprehend all this.
In this case, what would epsilon be representing specifically?
What is the name of these parts of calculus?
Also, my textbook says δ = 1/3 is the answer but how they got it I have no idea.