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Prove that limit as x approaches three of x^2 is equal to 9

  1. Aug 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that
    ##\lim_{x \rightarrow 3} x^{2} = 9##


    2. Relevant equations
    For every ε>0, there is a δ<0 so that if 0<|x-a|<δ then |f(x)-L|<ε

    3. The attempt at a solution
    ##If~0<|x - 3|<δ~then~|x^2 - 9|<ε##
    ##|x^2 - 9|##
    ##|x - 3||x + 3|##
    ##= |x - 3||x - 3 + 3 + 3|##
    ##= |x - 3|*|(x - 3) + 6|##
    ##≤ |x - 3|*(|x - 3| + |6|), by~triangle~inequality##
    ##= |x - 3|2 + 6|x - 3|##
    ##<##
     
  2. jcsd
  3. Aug 8, 2013 #2

    Zondrina

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    Homework Helper

    It was correct until there. Remember to use the fact that ##|x-3| < \delta## so you can write :

    ##|x - 3||x + 3| < \delta |x+3|##

    Now apply the triangle inequality to ##|x+3|##, what can you conclude?
     
  4. Aug 8, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    At this point, I would argue that if, say, |x- 3|< 1, then 2< x< 4 so that 5< x+ 3< 7.
    If we want [itex]|x- 3||x+ 3|< |x- 3|(7)< \epsilon[/itex] then we must have
    [itex]|x- 3|< \frac{\epsilon}{7}[/itex]
    So we can take [itex]\delta[/itex] to be the smaller of 1 and [itex]\frac{\epsilon}{7}[/itex]

     
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