Answer
$ a.\quad -\infty$
$ b.\quad +\infty$
$c.\quad 0$
$d.\displaystyle \quad\frac{3}{2}$
Work Step by Step
$a.$
When $x\rightarrow 0^{+}$z
the term $\displaystyle \frac{x^{2}}{2}\rightarrow 0$,
the term $(-\displaystyle \frac{1}{x})\rightarrow-\infty$ so,
$\displaystyle \lim_{x\rightarrow 0^{+}}(\frac{x^{2}}{2}-\frac{1}{x})=-\infty$
$b.$
When $x\rightarrow 0^{-}$
the term $\displaystyle \frac{x^{2}}{2}\rightarrow 0$,
the term $(-\displaystyle \frac{1}{x})\rightarrow-(-\infty)$ so,
$\displaystyle \lim_{x\rightarrow 0^{-}}(\frac{x^{2}}{2}-\frac{1}{x})=+\infty$
$c.$
Since $f(x)=\displaystyle \frac{x^{2}}{2}-\frac{1}{x}$ is continuous at $x= \sqrt[3]{2}$
$\displaystyle \lim_{x\rightarrow\sqrt[3]{2}}(\frac{x^{2}}{2}-\frac{1}{x})=\frac{\sqrt[3]{4}}{2}-\frac{1}{\sqrt[3]{2}}\cdot\frac{\sqrt[3]{}4}{\sqrt[3]{}4}=\frac{\sqrt[3]{4}}{2}-\frac{\sqrt[3]{4}}{2}=0$
$d.$
Since $f(x)=\displaystyle \frac{x^{2}}{2}-\frac{1}{x}$ is continuous at $x= -1$
$\displaystyle \lim_{x\rightarrow-1}(\frac{x^{2}}{2}-\frac{1}{x})=\frac{(-1)^{2}}{2}-\frac{1}{-1}=\frac{1}{2}+1=\frac{3}{2}$
