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Indeterminant limit (radical in denominator)

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{x \to {4}}\frac{4 - x^2}{2 - \sqrt{x}}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]\lim_{x \to {4}}\frac{4 - x^2}{2 - \sqrt{x}} \cdot \frac{2 + \sqrt{x}}{2 + \sqrt{x}}[/tex]

    [tex]= \frac{x(4-x)(2-\sqrt{x})}{(4-x)} = x(2-\sqrt{x}) [/tex]

    this equals zero, but is the limit indeterminate at this point?
     
    Last edited: Nov 19, 2007
  2. jcsd
  3. Nov 19, 2007 #2

    rock.freak667

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    Homework Helper

    I think you expanded the numerator incorrectly
     
  4. Nov 19, 2007 #3
    thanks! it was a missed minus sign.
     
  5. Feb 15, 2010 #4
    4-x^2 = (2-x)*(2+x)
     
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