Answer
$1.2$
Work Step by Step
Let us consider $f(x)=\lim\limits_{x \to -3}\dfrac{x^2-9}{x^2+x-6}$
Now, $x=-3.001 \implies f(x)=1.199960 \\x=-3.01 \implies f(x)= 1.199601 \\x=-3.1 \implies f(x)=1.196078$
From the computed data it has been seen that as $x$ approaches $-31$ from left and right, then $f(x)$ approaches $1.2$, so we have:
$ LHL =\lim\limits_{x \to 3^{-}}f(x)=1.2$ and $ RHL =\lim\limits_{x \to 3^{+}}f(x)=1.2$
Since, $ LHL =RHL$
Therefore, $\lim\limits_{x \to -3}f(x)=1.2$
