Answer
$$\frac{1}{6}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } \left( {3x + \sqrt {9{x^2} - x} } \right) \cr
& {\text{Rationalizing}} \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left[ {\frac{{3x + \sqrt {9{x^2} - x} }}{1} \times \frac{{3x - \sqrt {9{x^2} - x} }}{{3x - \sqrt {9{x^2} - x} }}} \right] \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{{\left( {3x} \right)}^2} - {{\left( {\sqrt {9{x^2} - x} } \right)}^2}}}{{3x - \sqrt {9{x^2} - x} }}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{9{x^2} - 9{x^2} + x}}{{3x - \sqrt {9{x^2} - x} }}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{x}{{3x - \sqrt {9{x^2} - x} }}} \right) \cr
& {\text{For }}x < 0,{\text{ we can write }}x = - \sqrt {{x^2}} .{\text{ So, dividing both the}} \cr
& {\text{numerator and denominator by }}x{\text{ produces}} \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{\frac{x}{x}}}{{\frac{{3x}}{x} - \frac{{\sqrt {9{x^2} - x} }}{{ - \sqrt {{x^2}} }}}}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{\frac{x}{x}}}{{\frac{{3x}}{x} + \sqrt {\frac{{9{x^2} - x}}{{{x^2}}}} }}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{1}{{3 + \sqrt {9 - \frac{1}{x}} }}} \right) \cr
& \cr
& {\text{Evaluate the limit}} \cr
& {\text{ = }}\frac{1}{{3 + \sqrt {9 - \frac{1}{{ - \infty }}} }} \cr
& {\text{ = }}\frac{1}{{3 + \sqrt 9 }} \cr
& = \frac{1}{6} \cr} $$
