Answer
$ - 9$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to - 3} \left( {2{x^3} + 6{x^2} - 9} \right) \cr
& {\text{Suppose that }}c{\text{ is a constant and the limits }} \cr
& {\text{ }}\mathop {\lim }\limits_{x \to a} f\left(x \right){\text{ and }}\mathop {\lim }\limits_{x \to a} g\left( x \right) \cr
& \cr
& {\text{Use the Sum and Difference Laws }} \cr
& \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right) \cr
& \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right) \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to - 3} \left( {2{x^3} + 6{x^2} - 9} \right) = \mathop {\lim }\limits_{x \to - 3} \left( {2{x^3}} \right) + \mathop {\lim }\limits_{x \to - 3} \left( {6{x^2}} \right) - \mathop {\lim }\limits_{x \to - 3} \left( 9 \right) \cr
& \cr
& {\text{Use the Constant Multiple Law }} \cr
& \mathop {\lim }\limits_{x \to a} \left[ {cf\left( x \right)} \right] = c\mathop {\lim }\limits_{x \to a} f\left( x \right) \cr
& = 2\mathop {\lim }\limits_{x \to - 3} \left( {{x^3}} \right) + 6\mathop {\lim }\limits_{x \to - 3} \left( {{x^2}} \right) - \mathop {\lim }\limits_{x \to - 3} \left( 9 \right) \cr
& \cr
& {\text{Use The power Law }}\mathop {\lim }\limits_{x \to a} {x^n} = {a^n}{\text{, }}n{\text{ is a positive integer }} \cr
& {\text{and the law}}\mathop {\lim }\limits_{x \to a} c = c,{\text{ then}} \cr
& = 2{\left( { - 3} \right)^3} + 6{\left( { - 3} \right)^2} - \left( 9 \right) \cr
& {\text{Simplifying}} \cr
& = 2\left( { - 27} \right) + 6\left( 9 \right) - \left( 9 \right) \cr
& = - 9 \cr} $$
