gostaria de saber, como solucionar a questão abaixo.

\[\eqalign{ & \mathop {\lim }\limits_{x \to - 3} \dfrac{{2{x^2} - 5x - 33}}{{{x^2} + 4x + 3}} = \mathop {\lim }\limits_{x \to - 3} \dfrac{{\left( {x + 3} \right)\left( {2x - 11} \right)}}{{\left( {x + 3} \right)\left( {x + 1} \right)}} \cr & \mathop {\lim }\limits_{x \to - 3} \dfrac{{2{x^2} - 5x - 33}}{{{x^2} + 4x + 3}} = \mathop {\lim }\limits_{x \to - 3} \dfrac{{2x - 11}}{{\left( {x + 1} \right)}} \cr & \mathop {\lim }\limits_{x \to - 3} \dfrac{{2{x^2} - 5x - 33}}{{{x^2} + 4x + 3}} = \mathop {\lim }\limits_{x \to - 3} \dfrac{{2\left( { - 3} \right) - 11}}{{\left( { - 3 + 1} \right)}} \cr & \mathop {\lim }\limits_{x \to - 3} \dfrac{{2{x^2} - 5x - 33}}{{{x^2} + 4x + 3}} = \dfrac{{ - 6 - 11}}{{ - 3 + 1}} \cr & \mathop {\lim }\limits_{x \to - 3} \dfrac{{2{x^2} - 5x - 33}}{{{x^2} + 4x + 3}} = \dfrac{{ - 17}}{{ - 2}} \cr & \mathop {\lim }\limits_{x \to - 3} \dfrac{{2{x^2} - 5x - 33}}{{{x^2} + 4x + 3}} = \dfrac{{17}}{2} }\]

Portanto, o resultado do limite dado será
\(\boxed{\dfrac{{17}}{2}}\)