Answer
$0$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{ - 2}}{{3x + 7}} \cr
& {\text{Using properties of limits}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{ - 2}}{{3x + 7}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( { - 2} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {3x + 7} \right)}} \cr
& {\text{ }} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( { - 2} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {3x} \right) + \mathop {\lim }\limits_{x \to \infty } \left( 7 \right)}} \cr
& {\text{Evaluate the limits}} \cr
& {\text{ }} = \frac{{ - 2}}{{\infty + 7}} \cr
& {\text{ }} = 0 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{ - 2}}{{3x + 7}} = 0 \cr} $$
