#### Answer

$$\frac{1}{6}$$

#### Work Step by Step

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