## Calculus with Applications (10th Edition)

$$\frac{1}{8}$$
\eqalign{ & \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {{x^2} + 7} - 4}}{{{x^2} - 9}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}3{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 3} \left( {\sqrt {{x^2} + 7} - 4} \right) = \sqrt {{3^2} + 7} - 4 = 0 \cr & \mathop {\lim }\limits_{x \to 3} \left( {{x^2} - 9} \right) = {2^2} - 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^2} + 7} - 4} \right) = \frac{{2x}}{{2\sqrt {{x^2} + 7} }} - 0 = \frac{x}{{\sqrt {{x^2} + 7} }} \cr & {\text{for }}{x^2} - 9 \to {D_x}\left( {{x^2} - 9} \right) = 2x \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {{x^2} + 7} - 4}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \frac{{\frac{x}{{\sqrt {{x^2} + 7} }}}}{{2x}} = \mathop {\lim }\limits_{x \to 3} \frac{1}{{2\sqrt {{x^2} + 7} }} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{{2\sqrt {{{\left( 3 \right)}^2} + 7} }} \cr & = \frac{1}{{2\sqrt {16} }} \cr & = \frac{1}{8} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {{x^2} + 7} - 4}}{{{x^2} - 9}} = \frac{1}{8} \cr}