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Math Help - finding limits graphically and numerically

  1. #1
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    finding limits graphically and numerically

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

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    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.
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    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.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by algebra2 View Post
    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?

    Quote Originally Posted by hwhelper View Post
    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
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    yeah, thanks for the explanation
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by algebra2 View Post
    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
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    See my signature for more details.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    See my signature for more details.
    your signature is so functional!
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