anonymous
  • anonymous
For the limit lim x → 2 (x3 − 4x + 3) = 3 illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1. (Round your answers to four decimal places.)
Calculus1
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anonymous
  • anonymous
For the limit lim x → 2 (x3 − 4x + 3) = 3 illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1. (Round your answers to four decimal places.)
Calculus1
schrodinger
  • schrodinger
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anonymous
  • anonymous
I know how to do parablas but not cubic
zepdrix
  • zepdrix
I don't remember these so well :) So take what I say with a grain of salt lol When your x is within delta of 2,\[\large\rm 0\lt|x-2|\lt\delta\]Then the distance between your function and the limiting value should be within epsilon,\[\large\rm |x^3-4x+3-3|\lt\epsilon\]So you've gotta do some work on this thing, ya?\[\large\rm |x^3-4x|\lt\epsilon\]\[\large\rm |x(x^2-4)|\lt\epsilon\]And then maybe apply Difference of Squares, ya? :o
zepdrix
  • zepdrix
\[\large\rm |x(x-2)(x+2)|\lt\epsilon\]And then umm... how do we wrap these up? I'm trying to remember haha

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anonymous
  • anonymous
umm your guess is better than mine lol. im teaching myself this
zepdrix
  • zepdrix
Let's put a restriction on delta, let it be small... say less than or equal to 1.\[\large\rm 0\lt|x-2|\lt\delta\le1\]\[\large\rm |x-2|\lt1\]\[\large\rm -1\lt x-2\lt 1\]\[\large\rm -2\lt x\lt 2\]So that ummm... If x is between 2 and -2, the junk in our epsilon equation is between these values,\[\large\rm 0\le|x(x+2)|\le8\]The larger of those is 8, so we'll use that.\[\large\rm |x(x+2)|~\cdot~|(x-2)|\lt\epsilon\]\[\large\rm |x(x+2)|~\cdot~|(x-2)|\lt8~\cdot~|(x-2)|\lt\epsilon\] So then uhhh... we would want the minimum value for this distance (x-2). Either \(\large\rm \frac{\epsilon}{8}\) or \(\large\rm \delta\).
zepdrix
  • zepdrix
Hopefully I'm doing this right :P Look at that, lemme know if it doesn't make sense..
anonymous
  • anonymous
Im confuse I think im going to a math lab thanks for your help
zepdrix
  • zepdrix
Woops, when I added the 2 to each side of the inequality it should have given this: \(\large\rm 1\lt x\lt 3\) Which would change things a bit. But ya, go to math lab c: figure some stuff out.

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