## Precalculus (6th Edition) Blitzer

Consider the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$. In making the table, choose the value of x close to $-2$ from the left and from the right as x approaches $-2$. As x approaches $-2$ from the left, arbitrarily start with $x=-2.01$. Then select two additional values of x that are closer to $-2$, but still less than $-2$; we choose $-2.001$ and $-2.0001$. Evaluate f at each chosen value of x to obtain the corresponding values of $f\left( x \right)$. At $x=-2.01$, Substitute, $x=-2.01$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$. Therefore, \begin{align} & \underset{x\to -2.01}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -2.01 \right)}^{3}}+8}{\left( -2.01 \right)+2} \\ & =\frac{-8.1206+8}{-0.01} \\ & =\frac{-0.1206}{-0.01} \\ & =12.0601 \end{align} Now, at $x=-2.001$ Substitute, $x=-2.001$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$. \begin{align} & \underset{x\to -2.001}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -2.001 \right)}^{3}}+8}{\left( -2.001 \right)+2} \\ & =\frac{-8.0120+8}{-0.001} \\ & =\frac{-0.0120}{-0.001} \\ & =12.0060 \end{align} At $x=-2.0001$ Substitute, $x=-2.0001$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$. \begin{align} & \underset{x\to -2.0001}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -2.0001 \right)}^{3}}+8}{\left( -2.0001 \right)+2} \\ & =\frac{-8.0012+8}{-0.0001} \\ & =\frac{-0.0012}{-0.0001} \\ & =12.0006 \end{align} Now, as x approaches $-2$ from the right, arbitrarily start with $x=-1.99$. Then select two additional values of x that are closer to $-2$, but still greater than $-2$; we choose $-1.999$ and $-1.9999$. At $x=-1.99$ Substitute, $x=-1.99$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$. Therefore, \begin{align} & \underset{x\to -1.99}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -1.99 \right)}^{3}}+8}{-1.99+2} \\ & =\frac{-7.8806+8}{0.01} \\ & =\frac{0.1194}{0.01} \\ & =11.9401 \end{align} Now, at $x=-1.999$ Substitute, $x=-1.999$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$. Therefore, \begin{align} & \underset{x\to -1.999}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -1.999 \right)}^{3}}+8}{-1.999+2} \\ & =\frac{-7.9880+8}{0.001} \\ & =\frac{0.012}{0.001} \\ & =11.9940 \end{align} At $x=-1.9999$ Substitute, $x=-1.9999$ in the provided limit notation, $\underset{x\to -2}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}$. \begin{align} & \underset{x\to -1.9999}{\mathop{\lim }}\,\frac{{{x}^{3}}+8}{x+2}=\frac{{{\left( -1.9999 \right)}^{3}}+8}{-1.9999+2} \\ & =\frac{-7.9988+8}{0.0001} \\ & =\frac{0.0012}{0.0001} \\ & =11.9994 \end{align} The limit of $\frac{{{x}^{3}}+8}{x+2}$ as x approaches $-2$ equals the number 12.