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Homework Help: Proving the limit (delta epsilon)

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

    Prove the follow statements directly using the formal [tex] \epsilon , \delta [/tex] definition.

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




    2. Relevant equations



    3. The attempt at a solution

    [tex] 0 < |x - 1| < \delta \rightarrow 0 < |\frac{x + 3}{x^2 + x + 4} - \frac{2}{3}| < \epsilon [/tex]

    Not sure what to do now.

    0 < | 3(x+3) - 2(x2 + x + 4) / 3(x2+x+4) | < e
    0 < | -2x2 + x + 1 / 3(x2+x+4) | < e
    0 < | (-2x - 1)(x - 1) / 3(x2+x+4) | < e

    Now I can control (x - 1), but how do I do this?
     
  2. jcsd
  3. Oct 9, 2009 #2
    Should I have posted this in the "Calculus and Beyond" section? :/
     
  4. Oct 9, 2009 #3

    HallsofIvy

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    Science Advisor

    Now you need to get a bound on [itex]|(2x+1)/3(x^2+ x+ 4)|[/itex]

    For example, you can start by assuming that |x-1|< 1 so that 0< x< 2. Then 0< 2x< 4 and 1< 2x+1< 5. Also 0< x2< 4 so 0< x2+ x< 6, 4< x2+ x+ 4< 10 and 12< 3(x2+ x+ 4)< 30. That tells you that 1/30< 1/3(x2+ x+ 4)< 1/12 and so that 1/30< (2x+1)/3(x2+ x+ 4)< 5/12.

    Now you know that |x-1|/30< (2x-1)|x-1|/3(x2+ x+ 4)< 5|x-1|/12

    Since you want to make sure that is less than [itex]\epsilon[/itex], you want [itex]5|x-1|/12<\epsilon[/itex] so you need [itex]|x-1|< 12\epsilon/5[/itex]. To make certain that |x-1|< 1 so all of that is true, take [itex]\delta[/itex] to be the smaller of 1 and [itex]12\epsilon/5|[/itex].
     
  5. Oct 14, 2009 #4
    I've been told that I should only pick values that are less than whatever x is approaching (ie. <1 in this case), is this true?
     
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