## 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

1. IrishBoy123

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 $$0 \lt |x-2| \lt \frac{\epsilon}{8}$$

2. anonymous

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

3. IrishBoy123

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...

4. IrishBoy123

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 $$\delta$$

5. anonymous

I think I understand it now, thanks

6. anonymous

Great work @IrishBoy123