Indeterminant limit (radical in denominator)

[furiouspoodle]

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?

 

[rock.freak667]

I think you expanded the numerator incorrectly

 

[furiouspoodle]

thanks! it was a missed minus sign.

 

[pmoseman]

4-x^2 = (2-x)*(2+x)