Answer
$ a.\quad+\infty$
$ b.\quad-\infty$
$ c.\quad-\infty$
$ d.\quad+\infty$
Work Step by Step
$a.$
When $x\rightarrow 2^{+}$, it is a positive number slightly greater than 2.
Its square is a number slightly more than 4.
The denominator is then slightly more than 0, which leads to
$\displaystyle \lim_{x\rightarrow 2^{+}} (\displaystyle \frac{1}{x^{2}-4})=+\infty$
$b.$
When $x\rightarrow 2^{-}$, it is a positive number slightly smaller than 2.
Its square is a number slightly less than 4.
The denominator is then slightly less than 0, which leads to
$\displaystyle \lim_{x\rightarrow 2^{-}} (\displaystyle \frac{1}{x^{2}-4})=-\infty$
$c.$
When $x\rightarrow-2^{+}$, it is a negative number whose absolute value is slightly smaller than 2.
Its square is a number slightly less than 4.
The denominator is then slightly less than 0, which leads to
$\displaystyle \lim_{x\rightarrow-2^{+}} (\displaystyle \frac{1}{x^{2}-4})=-\infty$
$d.$
When $x\rightarrow-2^{-}$, it is a negative number whose absolute value is slightly greater than 2.
Its square is a number slightly more than 4.
The denominator is then slightly more than 0, which leads to
$\displaystyle \lim_{x\rightarrow-2^{-}} (\displaystyle \frac{1}{x^{2}-4})=+\infty$
