Ta có $x\to (-3)^+\Rightarrow x+3>0$
$\begin{array}{l}
\mathop {\lim }\limits_{x \to - {3^ + }} \dfrac{{2 - x}}{{9 - {x^2}}} = \mathop {\lim }\limits_{x \to - {3^ + }} \dfrac{{2 - x}}{{3 - x}}.\mathop {\lim }\limits_{x \to - {3^ + }} \dfrac{1}{{x + 3}}\\
= \dfrac{{2 - \left( { - 3} \right)}}{{3 - \left( { - 3} \right)}}.\mathop {\lim }\limits_{x \to - {3^ + }} \dfrac{1}{{x + 3}} = + \infty
\end{array}$