## Calculus with Applications (10th Edition)

$$\frac{1}{8}$$
\eqalign{ & \mathop {\lim }\limits_{x \to 16} \frac{{\sqrt x - 4}}{{x - 16}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}16{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 16} \left( {\sqrt x - 4} \right) = \sqrt {16} - 4 = 0 \cr & \mathop {\lim }\limits_{x \to 16} \left( {x - 16} \right) = 16 - 16 = 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 - 4 - 3 \to {D_x}\left( {\sqrt x - 4} \right) = \frac{1}{{2\sqrt x }} - 0 \cr & {\text{for }}\left( {x - 16} \right) \to {D_x}\left( {x - 16} \right) = 1 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 16} \frac{{\sqrt x - 4}}{{x - 16}} = \mathop {\lim }\limits_{x \to 16} \frac{1}{{2\sqrt x }} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{{2\sqrt {16} }} \cr & = \frac{1}{{2\left( 4 \right)}} \cr & = \frac{1}{8} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 16} \frac{{\sqrt x - 4}}{{x - 16}} = \frac{1}{8} \cr}