I understand everything but the part in red. how do you get delta = E/3?

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- September 17th 2008, 03:56 PMalgebra2finding limits graphically and numerically
I understand everything but the part in red. how do you get delta = E/3?

http://img139.imageshack.us/img139/2568/helpte0.jpg - September 17th 2008, 04:04 PMhwhelperI think they mean "take delta = 1/3"
On the delta-epsilon formulation, delta is a free variable and you've to prove that for any positive delta you can find an epsilon such that abs(f(x) - limit) < epsilon. You could as well take delta = 1/2 or .0001.

- September 17th 2008, 04:08 PMhwhelper
My bad - read 1/3 in place of epsilon/3.

In any case I guess they mean you can take espilon = 3 * delta.

Point is you can choose any positive delta and that determines epsilon. - September 17th 2008, 04:28 PMJhevon
in fact, would work i think

anyway, recall what the definition of a limit means.

we want to find a , such that for all (and ), implies .

now, we found that , but we want , so we need to find a that works. now, what does it mean for to be close to 1? we give ourselves a generous range. and say, let it be somewhere between 0 and 2, and those x's are "close". now, to make sure our works, we decide to be cautious and choose , that way, gets really close to 1, since is the smallest for the range we are considering. so, if , we have , and so we choose that as our .

ok, so that was hopelessly confusing, even to me, i know what i want to say, but i am not sure if i said it ok. did you get that?

actually, it is delta that depends on epsilon, not the other way around. we must choose a delta that works for any epsilon we are given - September 17th 2008, 04:38 PMalgebra2
yeah, thanks for the explanation

- September 17th 2008, 04:52 PMJhevon
- September 17th 2008, 06:41 PMKrizalid
See my signature for more details.

- September 17th 2008, 06:42 PMJhevon