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Homework Help: Limit of a function with absolute value of polynomial in a quotient

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Lim | x2+x-12 |-8 / (x-4)
    x --> 4

    2. Relevant equations

    3. The attempt at a solution
    My answer is 9.
    It it right ?
    or there is not a limit for F(x) when x --> 4
  2. jcsd
  3. Dec 19, 2008 #2


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    Re: Limit

    Can we assume you mean [itex]\lim_{x\rightarrow 4} \left(\left|x^2+ x-12\right|- 8\right)/\left(x- 4\right)[/itex]?

    Close to 4, [itex]x^2+ x- 12[/itex] is close to +8 so the absolute value is not needed.
    [itex]x^2+ x- 12- 8= x^2+ x- 20= (x- 4)(x+ 5)[/itex] so [itex]\left(\left|x^2+ x-12\right|- 8\right)/\left(x- 4\right)= (x-4)(x+5)/(x-4)[/itex]. That, of course, has limit 4+ 5= 9 at x= 4.

    If, however, you meant
    [tex]|x^2+ x- 12|- \frac{8}{x- 4}[/tex]
    which what you actually wrote, that has no limit at x= 4.
  4. Dec 19, 2008 #3


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    Re: Limit

    Of course, without a calculation or proof, no answer is really right :P
    Did you sketch the graph, for instance? How can you see there whether there is a limit or not, as x --> 4?

    I think it is 9 too, actually.
  5. Dec 19, 2008 #4


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    Hi UNknown 2010! :smile:

    Yes, 9 is right :smile:

    I assume it's the | | that's worrying you?

    But it makes no difference at x = 4 (beacuse it's nowhere near 0 there).

    It would make a difference, and there would be no limit, if it were | x2+x-20 | / (x-4) :wink:
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