Answer
See below:
Work Step by Step
Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right),$ where, $ f\left( x \right)=\left\{ \begin{align}
& x+1\text{ if }x<0 \\
& 2x+1\ \text{ if }x\ge 0 \\
\end{align} \right\}$ .
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 $-0.0001$.
For $ x<0$, use the function $ f\left( x \right)=x+1$. Thus, the limit notation becomes $\underset{x\to 0}{\mathop{\lim }}\,x+1$.
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, $\underset{x\to 0}{\mathop{\lim }}\,x+1$.
Therefore,
$\begin{align}
& \underset{x\to -0.01}{\mathop{\lim }}\,x+1=-0.01+1 \\
& =0.99
\end{align}$
Now, at $ x=-0.001$
Substitute, $ x=-0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,x+1$.
$\begin{align}
& \underset{x\to -0.001}{\mathop{\lim }}\,x+1=-0.001+1 \\
& =0.999
\end{align}$
At $ x=-0.0001$
Substitute, $ x=-0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,x+1$.
$\begin{align}
& \underset{x\to -0.0001}{\mathop{\lim }}\,x+1=-0.0001+1 \\
& =0.9999
\end{align}$
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 $0.0001$.
For $ x\ge 0$, use the function $ f\left( x \right)=2x+1$. Thus, the limit notation becomes $\underset{x\to 0}{\mathop{\lim }}\,2x+1$.
At $ x=0.01$
Substitute, $ x=0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,2x+1$.
$\begin{align}
& \underset{x\to 0.01}{\mathop{\lim }}\,2x+1=2\left( 0.01 \right)+1 \\
& =0.02+1 \\
& =1.02
\end{align}$
At $ x=0.001$
Substitute, $ x=0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,2x+1$.
Therefore,
$\begin{align}
& \underset{x\to 0.001}{\mathop{\lim }}\,2x+1=2\left( 0.001 \right)+1 \\
& =0.002+1 \\
& =1.002
\end{align}$
At $ x=0.0001$
Substitute, $ x=0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,2x+1$.
$\begin{align}
& \underset{x\to 0.0001}{\mathop{\lim }}\,2x+1=2\left( 0.0001 \right)+1 \\
& =0.0002+1 \\
& =1.0002
\end{align}$
The limit of $ f\left( x \right)=\left\{ \begin{align}
& x+1\text{ if }x<0 \\
& 2x+1\ \text{ if }x\ge 0 \\
\end{align} \right\}$ as x approaches $0$ equals the number 1.