Chapter 11.1, Problem 14E

To determine

To find: The value of the $\underset{x\to -3}{\lim }\frac{x+3}{x^2+x-6}$ using table feature of a graphing utility. Also confirm your result graphically.

Answer to Problem 14E

The value of the limits is $\underset{x\to -3}{\lim }\frac{x+3}{x^2+x-6}=-0.2$

Explanation of Solution

Given information:

Given expression of limit as $\underset{x\to -3}{\lim }\frac{x+3}{x^2+x-6}$

Let, $f\left(x\right)=\frac{x+3}{x^2+x-6}$

The value of the function at $x=-3.1$ is,

  $\begin{array}{c}f\left(-3.1\right)=\frac{-3.1+3}{{\left(-3.1\right)}^2-3.1-6}\\ =\frac{-0.1}{9.61-9.1}\\ =\frac{-0.1}{0.51}\\ =-0.1960\end{array}$

The value of the function at $x=-3.01$ is,

  $\begin{array}{c}f\left(-3.01\right)=\frac{-3.01+3}{{\left(-3.01\right)}^2-3.01-6}\\ =\frac{-0.01}{9.0601-9.01}\\ =\frac{-0.01}{0.0501}\\ =-0.1996\end{array}$

The value of the function at $x=-3.001$ is,

  $\begin{array}{c}f\left(-3.001\right)=\frac{-3.001+3}{{\left(-3.001\right)}^2-3.001-6}\\ =\frac{-0.001}{9.006001-9.001}\\ =\frac{-0.001}{0.005001}\\ =-0.19996\end{array}$

The value of the function at $x=-3$ is,

  $\begin{array}{c}f\left(-3\right)=\frac{-3+3}{{\left(-3\right)}^2-3-6}\\ =\frac{0}{9-9}\\ =\frac{0}{0}\\ =\text{Undefined}\end{array}$

The value of the function at $x=-2.999$ is,

  $\begin{array}{c}f\left(-2.999\right)=\frac{-2.999+3}{{\left(-2.999\right)}^2-2.999-6}\\ =\frac{0.001}{8.994001-8.999}\\ =\frac{0.001}{-0.004999}\\ =-0.20\end{array}$

The value of the function at $x=-2.99$ is,

  $\begin{array}{c}f\left(-2.99\right)=\frac{-2.99+3}{{\left(-2.99\right)}^2-2.99-6}\\ =\frac{0.01}{8.9401-8.99}\\ =\frac{0.01}{-0.0499}\\ =-0.2004\end{array}$

The value of the function at $x=-2.9$ is,

  $\begin{array}{c}f\left(-2.9\right)=\frac{-2.9+3}{{\left(-2.9\right)}^2-2.9-6}\\ =\frac{0.1}{8.41-8.9}\\ =\frac{0.1}{-0.49}\\ =-0.2041\end{array}$

Hence, the complete table is,

  $\begin{array}{cccccccc}x& -3.1& -3.01& -3.001& -3& -2.999& -2.99& -2.9\\ f\left(x\right)& -0.1960& -0.1996& -0.19996& \text{undefined}& -0.20004& -0.2004& -0.2041\end{array}$

Therefore, the value of the limits is $\underset{x\to -3}{\lim }\frac{x+3}{x^2+x-6}=-0.2$

Graph of the function can be drawn as,

  

From the graph it is clear that $\underset{x\to -3}{\lim }\frac{x+3}{x^2+x-6}=-0.2$