Chapter 11 - Section 11.1 - Finding Limits Using Tables and Graphs - Exercise Set - Page 1138: 18

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Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right),$ where, $ f\left( x \right)=\left\{ \begin{align} & x+2\text{ if }x<0 \\ & 3x+2\ \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+2$. Thus, the limit notation becomes $\underset{x\to 0}{\mathop{\lim }}\,x+2$. 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, $\underset{x\to 0}{\mathop{\lim }}\,x+2$. Therefore, $\begin{align} & \underset{x\to -0.01}{\mathop{\lim }}\,x+2=-0.01+2 \\ & =1.99 \end{align}$ Now, at $ x=-0.001$ Substitute, $ x=-0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,x+2$. $\begin{align} & \underset{x\to -0.001}{\mathop{\lim }}\,x+2=-0.001+2 \\ & =1.999 \end{align}$ At $ x=-0.0001$ Substitute, $ x=-0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,x+2$. $\begin{align} & \underset{x\to -0.0001}{\mathop{\lim }}\,x+2=-0.0001+2 \\ & =1.9999 \end{align}$ 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)=3x+2$. Thus, the limit notation becomes $\underset{x\to 0}{\mathop{\lim }}\,3x+2$. At $ x=0.01$ Substitute, $ x=0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,3x+2$. $\begin{align} & \underset{x\to 0.01}{\mathop{\lim }}\,3x+2=3\left( 0.01 \right)+2 \\ & =0.03+2 \\ & =2.03 \end{align}$ At $ x=0.001$ Substitute, $ x=0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,3x+2$. Therefore, $\begin{align} & \underset{x\to 0.001}{\mathop{\lim }}\,3x+2=3\left( 0.001 \right)+2 \\ & =0.003+2 \\ & =2.003 \end{align}$ At $ x=0.0001$ Substitute, $ x=0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,3x+2$. $\begin{align} & \underset{x\to 0.0001}{\mathop{\lim }}\,3x+2=3\left( 0.0001 \right)+2 \\ & =0.0003+2 \\ & =2.0003 \end{align}$ The limit of $ f\left( x \right)=\left\{ \begin{align} & x+2\text{ if }x<0 \\ & 3x+2\ \text{ if }x\ge 0 \\ \end{align} \right\}$ as x approaches $0$ equals the number 2.