#### Answer

$${\text{ The limit does not exist}}$$

#### Work Step by Step

$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + 5x + 9} }}{{x - 1}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {{x^2} + 5x + 9} } \right) = \sqrt {{{\left( 1 \right)}^2} + 5\left( 1 \right) + 9} = \sqrt {15} \cr
& \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0 \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + 5x + 9} }}{{x - 1}} = \frac{{\sqrt {15} }}{0} \cr
& {\text{Since the limits do not leadsto the indeterminate form }}\frac{0}{0}{\text{ or }}\frac{{ \pm \infty }}{{ \pm \infty }} \cr
& {\text{we cannot apply the l'Hospital's rule}}. \cr
& {\text{By l'Hospital' rule}}{\text{, the limit does not exist}} \cr} $$