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Limit of Absolute Value

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data
    http://www4c.wolframalpha.com/Calculate/MSP/MSP621a026b7befdf1f1d00003b707661e1i20i4e?MSPStoreType=image/gif&s=55&w=126&h=38 [Broken]


    2. Relevant equations
    limit x->3- (3x-9)/((abs(3x-9))


    3. The attempt at a solution
    I don't understand how to find the limit of absolute values..
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 24, 2012 #2

    lanedance

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    what is the sign of |3x-9| for x<3?
     
  4. Jan 24, 2012 #3
    It would be -|3x-9| = -3x-9.. right?
     
  5. Jan 24, 2012 #4

    lanedance

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    the sign is negative - however what you have written is in correct

    teh way to do it is as follows
    x<3
    3x<9
    3x-9<0

    so it is negative, to make positive lets multiply by -1.

    So for x<3
    |3x-9| = -(3x-9) = 9-3x
     
  6. Jan 24, 2012 #5

    lanedance

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    however an even simpler way would be to do a subtitution (maybe, depending what you're comfortable with...)

    so you could let
    [tex] y=3x-9[/tex]

    then
    [tex] \lim_{x \to 3^-} \ \implies \ \lim_{y \to 0^-} [/tex]

    and the limit becomes
    [tex] \lim_{x \to 3^-} \frac{3x-9}{|3x-9|} \ \implies \ \lim_{y \to 0^-} \frac{y}{|y|} [/tex]
     
  7. Jan 24, 2012 #6
    Okay so the fraction is now 3x -9 / 9-3x? How do I get the answer of -1 though? Also, is the limit of y/abs(y) is always going to be -1?
     
  8. Jan 24, 2012 #7

    lanedance

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    1) how would you normally evaluate a limit, or where are you stuck. Is this one indeterminate?

    2) No. Maybe from the negative side, but that's what we're trying to show isn't it.

    Try out evaluating the direct limit first rather than the substitution, we can come back to that, but it is a helpful comparison
     
  9. Jan 24, 2012 #8
    Can you show me the steps to complete this problem because I have no idea where to start.
     
  10. Jan 25, 2012 #9

    lanedance

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    well let's star my considering the numerator and denominator separately, qualitatively, what happens to each x goes to 3 from the left

    the do you know about l'hopitals rule?
     
  11. Jan 25, 2012 #10
    We can't use L'Hopital's rule yet because my teacher hasn't taught it. Um, the numerator goes to infinity and so does the denominator. So infinity/infinity?
     
  12. Jan 25, 2012 #11

    lanedance

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    not quite, have another think, what happens to 3x-9 when x->3

    now this doesn't really need L'Hop as we have

    [tex] \lim_{x \to 3^-} \frac{3x-9}{|3x-9|} = \lim_{x \to 3^-} - \frac{3x-9}{3x-9}= - (\lim_{x \to 3^-} \frac{3x-9}{3x-9}) [/tex]

    as they're identical, i don't think it s a leap to say the final limit is -1
     
  13. Jan 25, 2012 #12
    Oh okay I think I finally understand. 3x-9 / 3x-9 is 1 and since there is a negative outside the parantheses, the limit is just -1 because you can't plug 3 into x since there is no x left.
     
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