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Prove the Limit

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Use the definition of limits to show that
    [tex]lim _{x\rightarrow 2} \frac{1}{x^{2}} = \frac{1}{4}[/tex]


    2. Relevant equations
    [tex]\forall \epsilon > 0, \exists \delta > 0, x \in D, 0 < |x-2| < \delta \Rightarrow |\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon [/tex]


    3. The attempt at a solution

    [tex]|\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon \Rightarrow |\frac{4-x^{2}}{4x^{2}}| < \epsilon \Rightarrow \frac{|4-x^{2}|}{4x^{2}} < \epsilon [/tex]

    [tex]Restrict: 1 < x < 3 \Rightarrow 4<4x^{2}<36 \Rightarrow \frac{1}{36} < \frac{1}{4x^{2}} < \frac{1}{4} \Rightarrow \frac{|4-x^{2}|}{36} < \frac{|4-x^{2}|}{4x^{2}} < \frac{|4-x^{2}|}{4} < \epsilon[/tex]

    And now I'm stuck. I'm trying to find delta based on epsilon. Normally the problems come to a neat little solution, but |4-x2| is not |x-2| and trying to get it to equal |x-2| is where I'm having the trouble.

    Any help is appreciated.
     
  2. jcsd
  3. Mar 19, 2009 #2

    gabbagabbahey

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    [itex]|4-x^2|=|x^2-4|=|(x-2)(x+2)|=[/itex]____?
     
  4. Mar 19, 2009 #3
    Yeah I got that far. I know that
    [tex]|(x-2)||(x+2)| < 4\epsilon[/tex]

    But that still leaves that (x+2) there (which I can remove the absolute value from since it has to be positive). I can't have delta dependent on x.

    I'm probably missing something blatant and obvious right here.
     
    Last edited: Mar 19, 2009
  5. Mar 19, 2009 #4

    gabbagabbahey

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    Well, you've restricted your domain to between 1 and 3, so..... __?__[itex]\leq|x+2|\leq[/itex]__?__ .....then simply choose epsilon to be greater than the upper bound of [tex]\frac{|x-2||x+2|}{4}[/tex]
     
  6. Mar 19, 2009 #5
    ah, duh, thanks. 3 < |x+2| < 5 (since x+2 is positive with or without the absolute value)

    So then

    [tex]\frac{|x-2||x+2|}{4} \leq \frac{5*|x-2|}{4} < \epsilon [/tex]

    Thus [tex]|x-2| < \frac{4}{5}\epsilon[/tex]

    Thus if [tex]\delta = \frac{4}{5}\epsilon \Rightarrow |f(x) - \frac{1}{4}| < \epsilon[/tex]


    Thanks so much.
     
  7. Mar 20, 2009 #6

    lurflurf

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    1<x<3
    implies
    3<x+2<5
     
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