Answer
See below:
Work Step by Step
Consider the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$ .
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=1.99$.
Then select two additional values of x that are closer to 2, but still less than 2; we choose 1.999 and 1.9999.
Evaluate f at each chosen value of x to obtain the corresponding values of $ f\left( x \right)$.
At $ x=1.99$,
Substitute, $ x=1.99$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$.
Therefore,
$\begin{align}
& \underset{x\to 1.99}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 1.99 \right)}^{2}} \\
& =5\left( 3.9601 \right) \\
& =19.801
\end{align}$
Now, at $ x=1.999$
Substitute, $ x=1.999$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$.
$\begin{align}
& \underset{x\to 1.999}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 1.999 \right)}^{2}} \\
& =5\left( 3.9960 \right) \\
& =19.980
\end{align}$
At $ x=1.9999$
Substitute, $ x=1.9999$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$.
$\begin{align}
& \underset{x\to 1.9999}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 1.9999 \right)}^{2}} \\
& =5\left( 3.9996 \right) \\
& =19.998
\end{align}$
Now, as x approaches 2 from the right, arbitrarily start with $ x=2.01$.
Then select two additional values of x that are closer to 2, but still greater than 2; we choose 2.001 and 2.0001.
At $ x=2.01$
Substitute, $ x=2.01$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$
$\begin{align}
& \underset{x\to 2.01}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 2.01 \right)}^{2}} \\
& =5\left( 4.0401 \right) \\
& =20.201
\end{align}$
Now, at $ x=2.001$
Substitute, $ x=2.001$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$
Therefore,
$\begin{align}
& \underset{x\to 2.001}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 2.001 \right)}^{2}} \\
& =5\left( 4.0040 \right) \\
& =20.020
\end{align}$
At $ x=2.0001$
Substitute, $ x=2.0001$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$
$\begin{align}
& \underset{x\to 2.0001}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 2.0001 \right)}^{2}} \\
& =5\left( 4.0004 \right) \\
& =20.002
\end{align}$
