anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Also, please explain the symbols.
anonymous
  • anonymous
they are just variable they use greek letters to annoy you
anonymous
  • anonymous
|dw:1348198061023:dw|

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

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