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Proving the limit #2

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \lim_{x\rightarrow 1} \frac{x + 3}{2x - 1} = 4
    [/tex]


    2. Relevant equations



    3. The attempt at a solution

    0 < |(x + 3 / 2x - 1) - 4| < e
    0 < | x+3-4(2x-1) / 2x-1 | < e
    0 < | -7x+7 / 2x-1 | < e
    0 < |-7(x-1) / 2x-1| < e

    Not sure what to do now.

    0 < ||-7||x-1|| / |2x-1| < e

    How do I relate the x-1 with d?
     
  2. jcsd
  3. Oct 9, 2009 #2

    tiny-tim

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    Hi zeion! :smile:
    oooh, that's so complicated! :cry:

    just write the original numerator as a multiple of 2x-1, plus a remainder …

    then it'll be obvious! :wink:
     
  4. Oct 9, 2009 #3
    You mean to write x+3 as a multiple of 2x-1?
     
  5. Oct 9, 2009 #4

    HallsofIvy

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    He means divide x+3 by 2x- 1: 2x+ 1 divides into x+ 3 "1/2" times with a remainder of 7/2:
    [tex]\frac{x+3}{2x-1}= \frac{1}{2}+ \frac{\frac{7}{2}}{2x-1}[/tex]
    though I honestly don't see how that simplifies a lot.

    Since you have already done the work, go ahead with
    [tex]0< 7\frac{|x-1|}{|2x-1|}< \epsilon[/tex]
    You just need a bound on 7/|2x-1|.

    Start by requiring that |x-1|< 1/4 so that -1/4< x- 1< 1/4 and 3/4< x< 5/4. Then 3/2< 2x< 5/2 so 1/2< 2x-1< 3/2. That tells you that 1/2< |2x-1|< 3/2.
    (I started with |x-1|< 1 but had to lower to 1/3 to keep those numbers larger than 0!)

    Now you know that 2/3< 1/|2x-1|< 2 and so that 7|x-1|/|2x-1|< 14|x-1|.
     
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