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anonymous
 one year ago
For the limit
lim
x → 2
(x^3 − 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.)
I know how to do it with a parabla but not cubic
anonymous
 one year ago
For the limit lim x → 2 (x^3 − 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.) I know how to do it with a parabla but not cubic

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.7you need \(\delta\) such that \(f(x)  3 < \epsilon\) for \(0 \lt x2 \lt \delta\) it is a cubic but an easy one \(x^3 − 4x + 3  3 \lt \epsilon\) \(x^3 − 4x \lt \epsilon\) \(x(x2)(x+2) \lt \epsilon\) as for the rest: \(x = 2 \pm \delta\) here so i think we can say \(2 \ x2 \ 4 \lt \epsilon\) \( x2 \lt \frac{\epsilon}{8}\) that might not be precise enough but you say you can do quadratics, so you might know better thus for \(0 \lt x2 \lt \delta\) we have \(0 \lt x2 \lt \frac{\epsilon}{8}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would you still need to subtract 2 since its subtract 2 in the middle?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.7if i understand you correctly, no. we are looking for a value for delta. i originally read your "I know how to do it with a parabla but not cubic" as suggesting the limits bit was fine, it was just playing with the cubic, but maybe i read wrong. we can say that for \(ε = 0.2 \ and \ 0.1\), \(\delta \lt 0.2/8 \ and \ \ 0.1/8\) are extremely good estimates [but i say estimates because the \(\frac{\epsilon}{8} \) is really \( \frac{\epsilon}{(2 \pm \delta)(4 \pm \delta) } \)] but they are asking for "the largest possible values of δ" which suggest that they want to push it to the limit???? and maybe a handwavey approach like this will be OK. i don't know. normally you might just assume that \(\epsilon\) is going to be small enough that \(\delta < 1\) is always true; so \(x2  \lt 1\), so \(1\lt x\lt 3\) and thus we can play safe with \(\delta = \frac{\epsilon}{x \ x+2} \lt \frac{\epsilon}{3 \times 5}\) i don't know the background, but it occurs to me that you should also test it. once you have your \(\delta\), plug \(x = 2 \pm \delta\) into the equation and see if the test we are trying to pass, ie \( f(x)−3<ϵ\), is true for those values. then you can fine tune and make sure you are correct. this is one way:p http://www.wolframalpha.com/input/?i=multiply+%282%2B0.9999k%2F8%29%28%2B0.9999k%2F8%29%284%2B0.9999k%2F8%29 let me know what you think...

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.7example 4 on this link, which starts half way down p3, better explains the usual kind of horsetrading or haircutting that you can do to justify limits in a delta epsilon proof, though in better detail that i have covered. *but* this is a long way away from finding the maximum value of \(\delta \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I understand it now, thanks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great work @IrishBoy123
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