Chapter 11, Problem 26RE

To determine

To calculate:

The limit value of the following functions with the help of graphing utility:

  $\underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}$

Answer to Problem 26RE

The limit value of $\underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}$ is $-\frac{1}{7}$ .

Explanation of Solution

Given information:

  $\underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}$

Calculation:

The given function is:

  $\underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}$

The limit of the function as $x\to -1$ is calculated as:

  $\begin{array}{l}\underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}=\underset{x\to -1}{\lim }\frac{x+1}{x^2-6x+x-6}\\ \underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}=\underset{x\to 5}{\lim }\frac{x+1}{x\left(x-6\right)+1\left(x-6\right)}\\ \underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}=\underset{x\to 5}{\lim }\frac{\left(x+1\right)}{\left(x-6\right)\left(x+1\right)}\\ \underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}=\underset{x\to 5}{\lim }\frac{1}{\left(x-6\right)}\\ \underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}=\frac{1}{\left(-1-6\right)}\left[\because x=-1\right]\\ \underset{x\to -1}{\lim }\frac{x+1}{x^2-5x-6}=-\frac{1}{7}\end{array}$

Hence, the limiting value is $\frac{1}{15}$ and its graph is drawn as below.