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Proving a limit

  1. May 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove using the epsilon-delta definition of a limit that 0 is the limit as x approaches 2 of x^2 - 4


    2. Relevant equations



    3. The attempt at a solution


    I've never actually done a limit proof like this before, so I just want to make sure that it's correct.

    [tex] |x^2 - 4| = |x-2||x+2| < \delta |x+2| [/tex].

    We can restrict the size of our delta-interval small enough so that for a fixed quantity c, [tex] \delta < c [/tex] and 0 <= 2 - c. Since [tex] \delta < c [/tex] then [tex] 2 - c < 2 - \delta [/tex] and [tex] 2 + \delta < 2 + c [/tex], and so [tex] 2 - c < x < 2 + c \Rightarrow |x + 2| < |4 + c| [/tex], and so [tex] \delta |x+2| < \delta |4 + c| = \epsilon [/tex] and taking [tex] \delta = \frac{\epsilon}{4+c} [/tex] shows that we can always find a suitable delta such that for each epsilon > 0, |x^2 - 4| < epsilon whenever |x-2| < delta

    How's this look?
     
  2. jcsd
  3. May 29, 2009 #2
    If you are going to restrict delta, you might as well pick a specific value. For instance, we can suppose delta is less than or equal to 1 so that |x-2| < 1 implying 1 < x < 3 so 3 < x + 2 < 5 and |x+2| < 5. Then we have d|x+2| < 5d. Letting d = min{1, e/5} works.

    Or, you can also note that |x+2| = |x -2 +4| =< |x-2| + 4 by the triangle inequality. Then you need d(d+4) = e, so you can use the quadratic formula to find delta as a function of epsilon.
     
  4. May 29, 2009 #3
    true, but you can assume c=1 (that is |x-2|<1 ) at the first place to simplify the proof. (so that |x-2||x+2|<5|x-2|<e, if |x-2|<e/5 )
     
  5. May 29, 2009 #4
    snipez, I like your second method of finding a suitable delta. Thanks!
     
  6. May 29, 2009 #5

    HallsofIvy

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    In order to assume |x-2|< 1 you must say [itex]\delta[/itex]= min(1, [itex]\epsilon[/itex]/5)- the smaller of the two numbers.
     
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