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Homework Help: Finding the largest value of delta greater than 0

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data
    For f(x) = sqrt(27-x), L=4, x_0 = 11 epsilon = 1, find the largest value of delta > 0 in the formal definition of a limit which ensures that |f(x) - L| < epsilon

    2. Relevant equations

    the formal def. of a limit:
    lim x->x_0 F(x) = L if, for every number epsilon > 0, there exists a corresponding number delta>0 such that
    0<|x-x_0| < delta implies |f(x) - L| < epsilon

    3. The attempt at a solution

    I've tried plugging multiple different things in to the equations, but i, sadly, have no idea what i'm doing.

    0 < |x-11| < epsilon and |sqrt(27-x) - 4| < 1.

    Attempting to solve the second equation i get x>2. If i plug that into the first equation i got -9. But that is definitely not the answer. I just don't know what all these numbers mean.
  2. jcsd
  3. Sep 10, 2012 #2


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    How about solving [itex]\displaystyle |\sqrt{27-x\ } - 4| = 1[/itex] for x, to start things off ?
  4. Sep 10, 2012 #3
    Do you mind if i walk you through my steps, if i make a mistake it should be easier to pinpoint the area :)

    So |sqrt(27-x)-4| < 1
    - add 4 to both sides

    |sqrt(27-x)| < 5
    -get rid of the square root

    |27-x|< 25
    -subtract 27 from each side

    |-x| < -2

    - and here is where i'm having my first problem. Do i multiply both sides by a negative allowing x to be a positive variable? Or do the absolute value signs take care of the negative variable?
  5. Sep 10, 2012 #4


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    Let me repeat,
    Solve [itex]\displaystyle |\sqrt{27-x\ } - 4| = 1[/itex] for x.​

    But if you insist, you can solve the inequality. Just be careful. That is trickier.

    At any rate:
    [itex]\displaystyle |\sqrt{27-x\ } - 4|+4\ne |\sqrt{27-x\ }|[/itex]​

    If [itex]\displaystyle |\sqrt{27-x\ } - 4| < 1[/itex]

    then [itex]\displaystyle -1<\sqrt{27-x\ } - 4 < 1\ .[/itex]

    Can you continue from there?
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