Answer
See below:
Work Step by Step
Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$.
In making the table, choose the value of x close to 0 from the left and from the right as x approaches 0.
As x approaches 0 from the left, arbitrarily start with $ x=-0.01$.
Then select two additional values of x that are closer to 0, but still less than 0; we choose $-0.001$ and $-0.0001$.
Evaluate f at each chosen value of x to obtain the corresponding values of $ f\left( x \right)$. All the values of x are in radians.
At $ x=-0.01$,
Substitute, $ x=-0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$.
Therefore,
$\underset{x\to -0.01}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( -0.01 \right)}^{2}}+\left( -0.01 \right)}{\sin \left( -0.01 \right)}$
Apply the trigonometric property, $\sin \left( -\theta \right)=-\sin \left( \theta \right)$
$\begin{align}
& \underset{x\to -0.01}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2\left( 0.0001 \right)-0.01}{-\sin \left( 0.01 \right)} \\
& =\frac{0.0002-0.01}{-0.0099} \\
& =\frac{-9.8\times {{10}^{-3}}}{-0.0099} \\
& =0.9800
\end{align}$
Now, at $ x=-0.001$
Substitute, $ x=-0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$.
$\underset{x\to -0.001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( -0.001 \right)}^{2}}+\left( -0.001 \right)}{\sin \left( -0.001 \right)}$
Apply the trigonometric property, $\sin \left( -\theta \right)=-\sin \left( \theta \right)$
Therefore,
$\begin{align}
& \underset{x\to -0.001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2\left( 1\times {{10}^{-6}} \right)-0.001}{-\sin \left( 0.001 \right)} \\
& =\frac{\left( 2\times {{10}^{-6}} \right)-0.001}{-\left( 9.99\times {{10}^{-4}} \right)} \\
& =\frac{-9.98\times {{10}^{-4}}}{-\left( 9.99\times {{10}^{-4}} \right)} \\
& =0.9980
\end{align}$
At $ x=-0.0001$
Substitute, $ x=-0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$.
$\underset{x\to -0.0001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( 0.0001 \right)}^{2}}+\left( -0.0001 \right)}{\sin \left( -0.0001 \right)}$
Apply the trigonometric property, $\sin \left( -\theta \right)=-\sin \left( \theta \right)$
$\begin{align}
& \underset{x\to -0.0001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2\left( 1\times {{10}^{-8}} \right)-0.0001}{-\sin \left( 0.0001 \right)} \\
& =\frac{\left( 2\times {{10}^{-8}} \right)-0.0001}{-\left( 9.99\times {{10}^{-5}} \right)} \\
& =\frac{-9.998\times {{10}^{-5}}}{-\left( 9.99\times {{10}^{-5}} \right)} \\
& =0.9998
\end{align}$
Now, as x approaches $0$ from the right, arbitrarily start with $ x=0.01$.
Then select two additional values of x that are closer to 0, but still greater than 0; we choose $0.001$ and $0.0001$.
At $ x=0.01$
Substitute, $ x=0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$.
Therefore,
$\underset{x\to 0.01}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( 0.01 \right)}^{2}}+\left( 0.01 \right)}{\sin \left( 0.01 \right)}$
$\begin{align}
& \underset{x\to 0.01}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2\left( 0.0001 \right)+0.01}{\sin \left( 0.01 \right)} \\
& =\frac{0.0002+0.01}{0.0099} \\
& =\frac{0.0102}{0.0099} \\
& =1.0200
\end{align}$
Now, at $ x=0.001$
Substitute, $ x=0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$
$\begin{align}
& \underset{x\to 0.001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( 0.001 \right)}^{2}}+\left( 0.001 \right)}{\sin \left( 0.001 \right)} \\
& =\frac{2\left( 1\times {{10}^{-6}} \right)+0.001}{9.99\times {{10}^{-4}}} \\
& =\frac{\left( 2\times {{10}^{-6}} \right)+0.001}{9.99\times {{10}^{-4}}} \\
& =\frac{1.002\times {{10}^{-3}}}{9.99\times {{10}^{-4}}}
\end{align}$
Further solve,
$\underset{x\to 0.001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=1.0020$
At $ x=0.0001$
Substitute, $ x=0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$.
Therefore,
$\begin{align}
& \underset{x\to 0.0001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( 0.0001 \right)}^{2}}+0.0001}{\sin \left( 0.0001 \right)} \\
& =\frac{2\left( 1\times {{10}^{-8}} \right)+0.0001}{9.99\times {{10}^{-5}}} \\
& =\frac{\left( 2\times {{10}^{-8}} \right)+0.0001}{\left( 9.99\times {{10}^{-5}} \right)} \\
& =\frac{1.0002\times {{10}^{-4}}}{\left( 9.99\times {{10}^{-5}} \right)} \\
& =1.0002
\end{align}$
The limit of $\frac{2{{x}^{2}}+x}{\sin x}$ as x approaches $0$ equals the number $1$.
