Answer
$ a.\quad +\infty$
$ b.\quad -\infty$
$c.\quad 0$
$d.\displaystyle \quad-\frac{1}{4}$
Work Step by Step
$a.$
When $x\rightarrow-2^{+}$, the absolute value of x is slightly less than 2, $x^{2}$ is a positive number, slightly less than 4 the numerator, $x^{2}-1$ is positive.
The denominator has (negative number, absolute value slightly below 4) +(4)
(the denominator is positive, close to zero).
$\displaystyle \lim_{x\rightarrow-2^{+}}(\frac{x^{2}-1}{2x+4})=+\infty$
$b.$
When $x\rightarrow-2^{-}$, the absolute value of x is slightly greater than 2, $x^{2}$ is a positive number, slightly greater than 4
the numerator, $x^{2}-1$ is positive.
The denominator has (negative number, absolute value slightly above 4) +(4)
(the denominator is negative, close to zero).
$\displaystyle \lim_{x\rightarrow-2^{-}}(\frac{x^{2}-1}{2x+4})=-\infty$
$c.$
Since $f(x)=\displaystyle \frac{x^{2}}{2}-\frac{1}{x}$ is continuous at $x= 1,$
$\displaystyle \lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow 1}f(x)=f(1)=\frac{1-1}{2+4}=0$
$d.$
Since $f(x)=\displaystyle \frac{x^{2}}{2}-\frac{1}{x}$ is continuous at $x= 0,$
$\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0}f(x)=f(1)=\frac{0-1}{0+4}=-\frac{1}{4}$
