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Limits of an Absolute Value

  1. Oct 10, 2011 #1
    Find the limit, if it exists. (If an answer does not exist, enter DNE.)
    lim x→−8 [ 8 − |x| OVER 8 + x ]






    The attempt at a solution

    First → abs(x) = -x ; x<0 / x ; x≥0

    Therefore, our equation is going to look like 8 − -x OVER 8 + x

    If approaching -8 from the left, I got 8 − -(-8-) OVER 8 + -8-

    → 0- OVER 0- . When dividing 0- over 0- you get infinity, correct?

    And doing the limit as x approaches -8 from the right gave me infinity as well, thus giving the answer, that as X approaches -8, Y approaches infinity. But that's wrong. Can someone help please?
     
  2. jcsd
  3. Oct 10, 2011 #2

    ehild

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    What is 8-(-x)???? Is not it the same as the denominator?

    ehild
     
  4. Oct 10, 2011 #3

    SammyS

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    0 over 0 is not infinity. It's an indeterminate form.

    (8-(-x))/(8+x) = ? provided that x ≠ -8
    (parentheses are important).
     
  5. Oct 10, 2011 #4

    HallsofIvy

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    The crucial point is that, for x going to -8, you can assume that x< 0. |x|= -x.
     
  6. Oct 10, 2011 #5
    Right, so I get the equation (8+x)/(8+x) = ? . Now I'm not entirely sure where to go from here.

    Just simply putting the value -8 into the equation gives 0/0 (which is wrong), and I'm not entirely sure how to change the way this equation looks..

    Would saying that it equals 1 be fair? Seeing as how the numerator and the denominator are the same?
     
  7. Oct 10, 2011 #6

    Mark44

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    Fair has nothing to do with it. Since the numerator and denominator are the same, for all negative values of x other than -8, the value of the expression (8 + x)/(8 + x) is 1. From this, you should be able to say what the value of the limit is.
     
  8. Oct 10, 2011 #7

    ehild

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    There is a definition, that a function f(x) has the limit A at x0 if to every sequence xn convergent to x0 the sequence f(xn) converges to A. This way a function can have a limit where it is not defined. Such an example is limx-->1f(x)=(x^2-1)/(x-1). f(x) is not defined at x=1, but for all x≠1 it is equal to x+1, so its limit is 2.

    ehild
     
  9. Oct 11, 2011 #8
    Ah thank you very much! And you are right, fair does have nothing to do with it.. I'll be sure to be more careful with my words.
     
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