Answer
See below:
Work Step by Step
Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$.
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=-1$.
Then select two additional values of x that are closer to 0, but still less than 0; we choose $-0.1$ and $-0.01$.
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=-1$,
Substitute, $ x=-1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$.
Therefore,
$\underset{x\to -1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( -1 \right)}{-1}$
Apply the trigonometric property, $\tan \left( -\theta \right)=-\tan \left( \theta \right)$.
$\begin{align}
& \underset{x\to -1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{-\tan \left( 1 \right)}{-1} \\
& =1.5574
\end{align}$
Now, at $ x=-0.1$
Substitute, $ x=-0.1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$.
$\underset{x\to -0.1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( -0.1 \right)}{-0.1}$
Apply the trigonometric property, $\tan \left( -\theta \right)=-\tan \left( \theta \right)$
$\begin{align}
& \underset{x\to -0.1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{-\tan \left( 0.1 \right)}{-\left( 0.1 \right)} \\
& =\frac{0.1003}{0.1} \\
& =1.0033
\end{align}$
At $ x=-0.01$
Substitute, $ x=-0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$.
$\underset{x\to -0.01}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( -0.01 \right)}{-\left( 0.01 \right)}$
Apply the trigonometric property, $\tan \left( -\theta \right)=-\tan \left( \theta \right)$
$\begin{align}
& \underset{x\to -0.01}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{-\tan \left( 0.01 \right)}{-\left( 0.01 \right)} \\
& =\frac{0.0100}{0.01} \\
& =1.00003
\end{align}$
Now, as x approaches $0$ from the right, arbitrarily start with $ x=1$.
Then select two additional values of x that are closer to $0$, but still greater than $0$; we choose $0.1$ and $0.01$.
At $ x=1$
Substitute, $ x=1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$.
Therefore,
$\begin{align}
& \underset{x\to 1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( 1 \right)}{1} \\
& =\frac{1.5574}{1} \\
& =1.5574
\end{align}$
Now, at $ x=0.1$
Substitute, $ x=0.1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$.
Therefore,
$\begin{align}
& \underset{x\to 0.1}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( 0.1 \right)}{0.1} \\
& =\frac{0.1003}{0.1} \\
& =1.0033
\end{align}$
At $ x=0.01$
Substitute, $ x=0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan x}{x}$.
$\begin{align}
& \underset{x\to 0.01}{\mathop{\lim }}\,\frac{\tan x}{x}=\frac{\tan \left( 0.01 \right)}{0.01} \\
& =\frac{0.0100}{0.01} \\
& =1.00003
\end{align}$
The limit of $\frac{\tan x}{x}$ as x approaches $0$ equals the number $1$.