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Math Help - Epsilon Delta proof, one side limit

  1. #1
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    [SOLVED]Epsilon Delta proof, one side limit

    Hey,
    I wanted to ask how can I proof the following statement using epsilon delta.
    lim x/[x] = 1
    x->3+

    *[x] means the whole part of x, for example : [3.5] = 3, [-3.5] = -4, [3] = 3.

    I do know how to proof normal statements using epsilon delta and so far I got to this point where I am stuck:
    |x/[x] -1| < e
    ...
    |x| < e/|1/[x] - 1/x|

    After that i decided to limit the neighborhood of 3 to a radius of 1, therefore:
    3 < x < 4
    ...
    0<1/[x] - 1/x < 1/12

    therefore:
    |x| < 12e

    Now I only have to choose an L so that if |x| < L, than |x| < 12e:
    3 < x < 3 + L
    But I cant find a proper L so that |x|< 12e.
    I thought using L = 12e - 3, and than :
    3 < x < 3 -3 +12e
    |x| < 12e
    But L must be larger than 0 for each e, so L = 12e -3 is no good.

    Thanks a lot,
    Michael.
    Last edited by MichaelEngstler; October 7th 2012 at 03:43 AM.
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  2. #2
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    Re: Epsilon Delta proof, one side limit

    I didn't follow parts of your work, but it seemed that you were making |x| small (<12e). The problem requires looking at x-3 small (positive). If you make x small (i.e. close to 0), then of course you'll have a hard time showing that x is also close to 3.

    \lim_{x \to 3^+} \frac{x}{[x]} = 1 means \forall \epsilon >0, \exists \delta>0 \ni 3<x<3+\delta \Rightarrow \left| \frac{x}{[x]} - 1 \right| < \epsilon

    The first step is, given an \epsilon>0, to find a \delta that works. The second step is to show that your discovered \delta does in fact work.

    1st) Find \delta:
    As always, we'll look at the epsilon expression of the limit, and try to manipulate it into looking like 3 < x < 3+something, where that "something" will become the delta.

    The first thing to note is that if \delta \le 1, then 3 < x < 3+\delta implies 3 < x < 4, so that [x] = 3.
    So, given \epsilon>0, we'll make sure that, whatever else we decide about \delta, we'll also have \delta \le1. This "gets rid of" that awkward function [x].

    Want \left| \frac{x}{[x]} - 1 \right| < \epsilon, so, since 3 < x < 4, want \left| \frac{x}{3} - 1 \right| < \epsilon.

    \left| \frac{x}{3} - 1 \right| < \epsilon iff 3\left| \frac{x}{3} - 1 \right| < 3\epsilon

    iff \left| (3)\left(\frac{x}{3}\right) - (3)(1) \right| < 3\epsilon iff |x- 3| < 3\epsilon.

    Now |x- 3| < 3\epsilon is the desired form. Thus want to set \delta = 3\epsilon. However, must also include the stipulation that \delta \le 1. So set \delta = min\{1, 3\epsilon\}.

    2nd) Prove that \delta works:

    Given any \epsilon > 0, let \delta = min\{1, 3\epsilon\}. Note that \delta > 0.

    Then 3 < x < 3+\delta implies 3 < x < 3+3\epsilon implies 0 < x-3 < 3\epsilon

    implies 0 < \frac{x-3}{3} < \epsilon implies 0 < \frac{x}{3} - 1 < \epsilon

    implies \left| \frac{x}{3} - 1 \right| < \epsilon.

    Also, since \delta \le 1 and 3 < x < 3+\delta, have that 3 < x < 4, so [x] = 3.

    Thus 3 < x < 3+\delta implies \left| \frac{x}{3} - 1 \right| < \epsilon implies \left| \frac{x}{[x]} - 1 \right| < \epsilon.

    This proves that, given any \epsilon > 0, if you choose \delta>0 by \delta = min\{1, 3\epsilon\}, then

    3 < x < 3+\delta implies \left| \frac{x}{[x]} - 1 \right| < \epsilon.

    That has proven that \lim_{x \to 3^+} \frac{x}{[x]} = 1.
    Last edited by johnsomeone; October 6th 2012 at 11:14 PM.
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  3. #3
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    Re: Epsilon Delta proof, one side limit

    Thanks a lot, you are awesome !
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