# finding limits graphically and numerically

• Sep 17th 2008, 04:56 PM
algebra2
finding 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
• Sep 17th 2008, 05:04 PM
hwhelper
I 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.
• Sep 17th 2008, 05:08 PM
hwhelper
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.
• Sep 17th 2008, 05:28 PM
Jhevon
Quote:

Originally Posted by algebra2
I understand everything but the part in red. how do you get delta = E/3?

http://img139.imageshack.us/img139/2568/helpte0.jpg

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?

Quote:

Originally Posted by hwhelper
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
• Sep 17th 2008, 05:38 PM
algebra2
yeah, thanks for the explanation
• Sep 17th 2008, 05:52 PM
Jhevon
Quote:

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
• Sep 17th 2008, 07:41 PM
Krizalid
See my signature for more details.
• Sep 17th 2008, 07:42 PM
Jhevon
Quote:

Originally Posted by Krizalid
See my signature for more details.

your signature is so functional! :D