## Calculus with Applications (10th Edition)

$$\frac{1}{6}$$
\eqalign{ & \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}9{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 9} \left( {\sqrt x - 3} \right) = \sqrt 9 - 3 = 0 \cr & \mathop {\lim }\limits_{x \to 9} \left( {x - 9} \right) = 9 - 9 = 0 \cr & \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr & {\text{denominator}}{\text{}} \cr & {\text{for }}\sqrt x - 3 - 3 \to {D_x}\left( {\sqrt x - 3} \right) = \frac{1}{{2\sqrt x }} - 0 \cr & {\text{for }}\left( {x - 9} \right) \to {D_x}\left( {x - 9} \right) = 1 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} = \mathop {\lim }\limits_{x \to 9} \frac{1}{{2\sqrt x }} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{{2\sqrt 9 }} \cr & = \frac{1}{{2\left( 3 \right)}} \cr & = \frac{1}{6} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} = \frac{1}{6} \cr}