finding limits graphically and numerically

**algebra2** · September 17th 2008, 03:56 PM

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

**hwhelper** · September 17th 2008, 04:04 PM

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.

**hwhelper** · September 17th 2008, 04:08 PM

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.

**Jhevon** · September 17th 2008, 04:28 PM

Originally Posted by algebra2

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

in fact, $\delta = \frac {\epsilon}2$ would work i think

anyway, recall what the definition of a limit means.

we want to find a $\delta > 0$ , such that for all $\epsilon > 0$ (and $x \in \text{dom}(f)$ ), $|x - 1| < \delta$ implies $|f(x) - 2| < \epsilon$ .

now, we found that $|x - 1| < \frac {\epsilon}{x + 1}$ , but we want $|x - 1|< \delta$ , so we need to find a $\delta$ that works. now, what does it mean for $x$ 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 $\delta$ works, we decide to be cautious and choose $x = 2$ , that way, $x$ gets really close to 1, since $\frac {\epsilon}{x + 1}$ is the smallest for the range we are considering. so, if $x = 2$ , we have $\frac {\epsilon}{x + 1} = \frac {\epsilon}3$ , and so we choose that as our $\delta$ .

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?

Originally Posted by hwhelper

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.

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

**algebra2** · September 17th 2008, 04:38 PM

yeah, thanks for the explanation

**Jhevon** · September 17th 2008, 04:52 PM

Originally Posted by algebra2

yeah, thanks for the explanation

good, i was worried. i hope you realize that things like $|x - 1|< \delta$ is talking about distance. the distance between x and 1 is less than delta, that's what it means. this is why i was talking about x "close" to 1, etc

**Krizalid** · September 17th 2008, 06:41 PM

See my signature for more details.

**Jhevon** · September 17th 2008, 06:42 PM

Originally Posted by Krizalid

See my signature for more details.

your signature is so functional!

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finding limits graphically and numerically

I think they mean "take delta = 1/3"

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