Homework Help: Indeterminant limit (radical in denominator)

1. Nov 19, 2007

furiouspoodle

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

Last edited: Nov 19, 2007
2. Nov 19, 2007

rock.freak667

I think you expanded the numerator incorrectly

3. Nov 19, 2007

furiouspoodle

thanks! it was a missed minus sign.

4. Feb 15, 2010

pmoseman

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