## Thomas' Calculus 13th Edition

$a.\quad +\infty$ $b.\quad -\infty$ $c.\quad 0$ $d.\displaystyle \quad-\frac{1}{4}$
$a.$ 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 $x= 0,$ $\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0}f(x)=f(1)=\frac{0-1}{0+4}=-\frac{1}{4}$