Answer
$$\frac{1}{{12}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 5} \frac{{\sqrt {{x^2} + 11} - 6}}{{{x^2} - 25}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}5{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 5} \left( {\sqrt {{x^2} + 11} - 6} \right) = \sqrt {{5^2} + 11} - 6 = 6 - 6 = 0 \cr
& \mathop {\lim }\limits_{x \to 5} \left( {{x^2} - 25} \right) = {5^2} - 25 = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{}} \cr
& {\text{for }}\sqrt {{x^2} + 11} - 6 \to {D_x}\left( {\sqrt {{x^2} + 11} - 6} \right) = \frac{{2x}}{{2\sqrt {{x^2} + 11} }} - 0 = \frac{x}{{\sqrt {{x^2} + 11} }} \cr
& {\text{for }}{x^2} - 25 \to {D_x}\left( {{x^2} - 25} \right) = 2x \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 5} \frac{{\sqrt {{x^2} + 11} - 6}}{{{x^2} - 25}} = \mathop {\lim }\limits_{x \to 5} \frac{{\frac{x}{{\sqrt {{x^2} + 11} }}}}{{2x}} = \mathop {\lim }\limits_{x \to 5} \frac{1}{{2\sqrt {{x^2} + 11} }} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{1}{{2\sqrt {{{\left( 5 \right)}^2} + 11} }} \cr
& = \frac{1}{{2\sqrt {36} }} \cr
& = \frac{1}{{12}} \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 5} \frac{{\sqrt {{x^2} + 11} - 6}}{{{x^2} - 25}} = \frac{1}{{12}} \cr} $$
