Answer
See below:
Work Step by Step
Consider the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$.
In making the table, choose the value of x close to $-2$ from the left and from the right as x approaches $-2$.
As x approaches $-2$ from the left, arbitrarily start with $ x=-2.01$.
Then select two additional values of x that are closer to $-2$, but still less than $-2$; we choose $-2.001$ and $-2.0001$.
Evaluate f at each chosen value of x to obtain the corresponding values of $ f\left( x \right)$.
At $ x=-2.01$,
Substitute, $ x=-2.01$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$.
Therefore,
$\begin{align}
& \underset{x\to -2.01}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -2.01 \right)}^{3}}+8}{\left( -2.01 \right)+2} \\
& =\frac{-8.1206+8}{-0.01} \\
& =\frac{-0.1206}{-0.01} \\
& =12.0601
\end{align}$
Now, at $ x=-2.001$
Substitute, $ x=-2.001$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$.
$\begin{align}
& \underset{x\to -2.001}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -2.001 \right)}^{3}}+8}{\left( -2.001 \right)+2} \\
& =\frac{-8.0120+8}{-0.001} \\
& =\frac{-0.0120}{-0.001} \\
& =12.0060
\end{align}$
At $ x=-2.0001$
Substitute, $ x=-2.0001$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$.
$\begin{align}
& \underset{x\to -2.0001}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -2.0001 \right)}^{3}}+8}{\left( -2.0001 \right)+2} \\
& =\frac{-8.0012+8}{-0.0001} \\
& =\frac{-0.0012}{-0.0001} \\
& =12.0006
\end{align}$
Now, as x approaches $-2$ from the right, arbitrarily start with $ x=-1.99$.
Then select two additional values of x that are closer to $-2$, but still greater than $-2$; we choose $-1.999$ and $-1.9999$.
At $ x=-1.99$
Substitute, $ x=-1.99$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$.
Therefore,
$\begin{align}
& \underset{x\to -1.99}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -1.99 \right)}^{3}}+8}{-1.99+2} \\
& =\frac{-7.8806+8}{0.01} \\
& =\frac{0.1194}{0.01} \\
& =11.9401
\end{align}$
Now, at $ x=-1.999$
Substitute, $ x=-1.999$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$.
Therefore,
$\begin{align}
& \underset{x\to -1.999}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -1.999 \right)}^{3}}+8}{-1.999+2} \\
& =\frac{-7.9880+8}{0.001} \\
& =\frac{0.012}{0.001} \\
& =11.9940
\end{align}$
At $ x=-1.9999$
Substitute, $ x=-1.9999$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$.
$\begin{align}
& \underset{x\to -1.9999}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -1.9999 \right)}^{3}}+8}{-1.9999+2} \\
& =\frac{-7.9988+8}{0.0001} \\
& =\frac{0.0012}{0.0001} \\
& =11.9994
\end{align}$
The limit of $\frac{{{x}^{3}}+8}{x+2}$ as x approaches $-2$ equals the number 12.
