Precalculus (6th Edition) Blitzer

Consider the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$. In making the table, choose the value of x close to $-5$ from the left and from the right as x approaches $-5$. As x approaches $-5$ from the left, arbitrarily start with $x=-5.1$. Then select two additional values of x that are closer to $-5$, but still less than $-5$; we choose $-5.01$ and $-5.001$. Evaluate f at each chosen value of x to obtain the corresponding values of $f\left( x \right)$. At $x=-5.1$, Substitute, $x=-5.1$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$. Therefore, \begin{align} & \underset{x\to -5.1}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -5.1 \right)}^{2}}-25}{\left( -5.1 \right)+5} \\ & =\frac{26.01-25}{-0.1} \\ & =\frac{1.01}{-0.1} \\ & =-10.1 \end{align} Now, at $x=-5.01$ Substitute, $x=-5.01$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$. \begin{align} & \underset{x\to -5.01}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -5.01 \right)}^{2}}-25}{\left( -5.01 \right)+5} \\ & =\frac{25.1001-25}{-0.01} \\ & =\frac{0.1001}{-0.01} \\ & =-10.01 \end{align} At $x=-5.001$ Substitute, $x=-5.001$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$. \begin{align} & \underset{x\to -5.001}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -5.001 \right)}^{2}}-25}{\left( -5.001 \right)+5} \\ & =\frac{25.0100-25}{-0.001} \\ & =\frac{0.001}{-0.01} \\ & =-10.001 \end{align} Now, as x approaches $-5$ from the right, arbitrarily start with $x=-4.9$. Then select two additional values of x that are closer to $-5$, but still greater than $-5$; we choose $-4.99$ and $-4.999$. At $x=-4.9$ Substitute, $x=-4.9$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$. Therefore, \begin{align} & \underset{x\to -4.9}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -4.9 \right)}^{2}}-25}{-4.9+5} \\ & =\frac{24.01-25}{0.1} \\ & =\frac{-0.99}{0.1} \\ & =-9.9 \end{align} Now, at $x=-4.99$ Substitute, $x=-4.99$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$. Therefore, \begin{align} & \underset{x\to -4.99}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -4.99 \right)}^{2}}-25}{-4.99+5} \\ & =\frac{24.9001-25}{0.01} \\ & =\frac{-0.0999}{0.01} \\ & =-9.99 \end{align} At $x=-4.999$ Substitute, $x=-4.999$ in the provided limit notation, $\underset{x\to -5}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}$. \begin{align} & \underset{x\to -4.999}{\mathop{\lim }}\,\frac{{{x}^{2}}-25}{x+5}=\frac{{{\left( -4.999 \right)}^{2}}-25}{-4.999+5} \\ & =\frac{24.9900-25}{0.001} \\ & =\frac{-0.01}{0.001} \\ & =-9.999 \end{align} The limit of $\frac{{{x}^{3}}+8}{x+2}$ as x approaches $-5$ equals the number $-10$.