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