Indeterminant limit (radical in denominator)

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SUMMARY

The limit of the expression \(\lim_{x \to 4}\frac{4 - x^2}{2 - \sqrt{x}}\) is evaluated using algebraic manipulation. By multiplying the numerator and denominator by the conjugate \(2 + \sqrt{x}\), the expression simplifies to \(x(2 - \sqrt{x})\). At \(x = 4\), this limit evaluates to zero, confirming that the limit is not indeterminate. The initial confusion stemmed from an incorrect expansion of the numerator, which should be expressed as \(4 - x^2 = (2 - x)(2 + x)\).

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Homework Statement



\lim_{x \to {4}}\frac{4 - x^2}{2 - \sqrt{x}}

Homework Equations


The Attempt at a Solution



\lim_{x \to {4}}\frac{4 - x^2}{2 - \sqrt{x}} \cdot \frac{2 + \sqrt{x}}{2 + \sqrt{x}}

= \frac{x(4-x)(2-\sqrt{x})}{(4-x)} = x(2-\sqrt{x})

this equals zero, but is the limit indeterminate at this point?
 
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I think you expanded the numerator incorrectly
 
thanks! it was a missed minus sign.
 
4-x^2 = (2-x)*(2+x)
 

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