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Indeterminant limit (radical in denominator)

  • #1

Homework Statement



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

Homework Equations





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:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
I think you expanded the numerator incorrectly
 
  • #3
thanks! it was a missed minus sign.
 
  • #4
14
0
4-x^2 = (2-x)*(2+x)
 

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