Answer
$0$
Work Step by Step
Let us consider $f(x)=\lim\limits_{x \to 2}\dfrac{x^2-4}{x+2}$
Now, $x=2.001 \implies f(x)= 0.001000; \\x=2.01 \implies f(x)= 0.01000 \\x=2.1 \implies f(x)=0.100000$
From the computed data it has been seen that as $x$ approaches $2$ from left and right, then $f(x)$ approaches $6$, so we have:
$ LHL =\lim\limits_{x \to 2^{-}}f(x)=0$ and $ RHL =\lim\limits_{x \to 2^{+}}f(x)=0$
Since, $ LHL =RHL$
Therefore, $\lim\limits_{x \to 2}f(x)=0$
