Answer
See below:
Work Step by Step
Consider the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$.
In making the table, choose the value of x close to $-5$ from the left and from the right as x approaches $-5$.
As x approaches $-5$ from the left, arbitrarily start with $ x=-5.1$.
Then select two additional values of x that are closer to $-5$, but still less than $-5$; we choose $-5.01$ and $-5.001$.
Evaluate f at each chosen value of x to obtain the corresponding values of $ f\left( x \right)$.
At $ x=-5.1$,
Substitute, $ x=-5.1$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$.
Therefore,
$\begin{align}
& \underset{x\to -5.1}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -5.1 \right)}^{2}}-25}{\left( -5.1 \right)+5} \\
& =\frac{26.01-25}{-0.1} \\
& =\frac{1.01}{-0.1} \\
& =-10.1
\end{align}$
Now, at $ x=-5.01$
Substitute, $ x=-5.01$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$.
$\begin{align}
& \underset{x\to -5.01}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -5.01 \right)}^{2}}-25}{\left( -5.01 \right)+5} \\
& =\frac{25.1001-25}{-0.01} \\
& =\frac{0.1001}{-0.01} \\
& =-10.01
\end{align}$
At $ x=-5.001$
Substitute, $ x=-5.001$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$.
$\begin{align}
& \underset{x\to -5.001}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -5.001 \right)}^{2}}-25}{\left( -5.001 \right)+5} \\
& =\frac{25.0100-25}{-0.001} \\
& =\frac{0.001}{-0.01} \\
& =-10.001
\end{align}$
Now, as x approaches $-5$ from the right, arbitrarily start with $ x=-4.9$.
Then select two additional values of x that are closer to $-5$, but still greater than $-5$; we choose $-4.99$ and $-4.999$.
At $ x=-4.9$
Substitute, $ x=-4.9$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$.
Therefore,
$\begin{align}
& \underset{x\to -4.9}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -4.9 \right)}^{2}}-25}{-4.9+5} \\
& =\frac{24.01-25}{0.1} \\
& =\frac{-0.99}{0.1} \\
& =-9.9
\end{align}$
Now, at $ x=-4.99$
Substitute, $ x=-4.99$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$.
Therefore,
$\begin{align}
& \underset{x\to -4.99}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -4.99 \right)}^{2}}-25}{-4.99+5} \\
& =\frac{24.9001-25}{0.01} \\
& =\frac{-0.0999}{0.01} \\
& =-9.99
\end{align}$
At $ x=-4.999$
Substitute, $ x=-4.999$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$.
$\begin{align}
& \underset{x\to -4.999}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -4.999 \right)}^{2}}-25}{-4.999+5} \\
& =\frac{24.9900-25}{0.001} \\
& =\frac{-0.01}{0.001} \\
& =-9.999
\end{align}$
The limit of $\frac{{{x}^{3}}+8}{x+2}$ as x approaches $-5$ equals the number $-10$.
