## Precalculus (6th Edition) Blitzer

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.