Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } \left( {x + \sqrt {{x^2} + 3} } \right) \cr
& {\text{Rationalizing}} \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left[ {\frac{{x + \sqrt {{x^2} + 3} }}{1} \times \frac{{x - \sqrt {{x^2} + 3} }}{{x - \sqrt {{x^2} + 3} }}} \right] \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{x^2} - {{\left( {\sqrt {{x^2} + 3} } \right)}^2}}}{{x + \sqrt {{x^2} + 3} }}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{x^2} - {x^2} - 3}}{{x + \sqrt {{x^2} + 3} }}} \right) \cr
& {\text{ = }} - \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{3}{{x + \sqrt {{x^2} + 3} }}} \right) \cr
& {\text{Evaluate the limit}} \cr
& {\text{ = }} - \left( {\frac{3}{{ - \infty + \sqrt {{{\left( { - \infty } \right)}^2} + 3} }}} \right) \cr
& {\text{ = }} - \frac{3}{{ - \infty }} \cr
& = 0 \cr} $$
