## Calculus, 10th Edition (Anton)

$\lim\limits_{x \to 9}\frac{x-9}{\sqrt x - 3}$ = 6
($\sqrt x$ - 3)($\sqrt x$ + 3) = x - 9. So: $\frac{x-9}{\sqrt x - 3}$$\times$$\frac{\sqrt x+3}{\sqrt x+3}$ = $\frac{(x-9)(\sqrt x+3)}{x-9}$ = $\sqrt x$ + 3. Now we have: $\lim\limits_{x \to 9}\frac{x-9}{\sqrt x - 3}$ = $\lim\limits_{x \to 9}\sqrt x + 3$ = $\sqrt 9$ + 3 = 6