Question

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

Answer 1

you need \(\delta\) such that \(|f(x) - 3| < \epsilon\) for \(0 \lt |x-2| \lt \delta\)

it is a cubic but an easy one
\(|x^3 − 4x + 3 - 3| \lt \epsilon\)
\(|x^3 − 4x| \lt \epsilon\)
\(|x(x-2)(x+2)| \lt \epsilon\)

as for the rest:
\(x = 2 \pm \delta\) here so i think we can say
\(|2| \ |x-2| \ |4| \lt \epsilon\)
\( |x-2| \lt \frac{\epsilon}{8}\)

that might not be precise enough but you say you can do quadratics, so you might know better

thus for \(0 \lt |x-2| \lt \delta\) we have \(0 \lt |x-2| \lt \frac{\epsilon}{8}\)

Answer 2

Would you still need to subtract 2 since its subtract 2 in the middle?

Answer 3

if 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 hand-wavey 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 \(|x-2 | \lt 1\), 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...

Answer 4

example 4 on this link, which starts half way down p3, better explains the usual kind of horse-trading 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 \(\delta \)

Answer 5

I think I understand it now, thanks

Answer 6

Great work @IrishBoy123