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Absolute value limit

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  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4}##.

    2. Relevant equations


    3. The attempt at a solution
    I am trying to solve the following limit: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4} = \lim_{x\rightarrow 2} \frac{\left | (x + 1)(x + 2) \right |}{(x - 2)(x + 2)}##. However, I am not sure how to proceed. Normally, I would cancel out the factors, but I am not sure what do with the absolute value in the numerator.
     
  2. jcsd
  3. Oct 23, 2015 #2

    Mark44

    Staff: Mentor

    Since x is near 2, |x + 2| will be near 4; i.e., positive.
     
  4. Oct 23, 2015 #3
    So I can just proceed as if there were no absolute value?
     
  5. Oct 23, 2015 #4

    PeroK

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    Are you sure it's not supposed to be ##x^2 - 3x + 2## in the numerator?
     
  6. Oct 23, 2015 #5
    Pretty sure. Why?
     
  7. Oct 23, 2015 #6

    PeroK

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    See whether you can work it out!
     
  8. Oct 23, 2015 #7
    Well I got DNE! Isn't that still a valid response?
     
  9. Oct 23, 2015 #8

    PeroK

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    For the original limit, assuming DNE means "does not exist", that's correct.

    ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 - 3x + 2 \right |}{x^2 - 4}##.

    Is a better problem, because it's of the form ##\frac{0}{0}## which the original problem wasn't.
     
  10. Oct 24, 2015 #9

    HallsofIvy

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    The point is that as x goes to 2, the denominator goes to 0 but the numerator does NOT. For x arbitrarily close to 0, the numerator close to 18, denominator close to 0, the fraction will be huge. If the limit were as x goes to -2, then it would be of the form "0/0".
     
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