Chapter 11.3, Problem 11E

Solution

a)

To explain: Why you cannot use substitution to find the limit $\underset{x\to -3}{\lim }\frac{x^2+7x+12}{x^2-9}$ .

When one substitutes -3 for $x$ then the denominator becomes zero, so one cannot use substitution to find the limit $\underset{x\to -3}{\lim }\frac{x^2+7x+12}{x^2-9}$ .

Given information: The given limit is $\underset{x\to -3}{\lim }\frac{x^2+7x+12}{x^2-9}$ .

Calculation:

Substitute $-3$ for $x$ in the given limit then the denominator becomes zero as shown below:

  $\begin{array}{c}{x}^2-9=3^2-9\\ =9-9\\ =0\end{array}$

Thus, the denominator becomes zero so one cannot use substitution to find the given limit.

b)

To find: The limit $\underset{x\to -3}{\lim }\frac{x^2+7x+12}{x^2-9}$ algebraically if it exists.

The limit $\underset{x\to -3}{\lim }\frac{x^2+7x+12}{x^2-9}$ exists and is found by factoring numerator and denominator as $-\frac{1}{6}$ .

Given information: The given limit is $\underset{x\to -3}{\lim }\frac{x^2+7x+12}{x^2-9}$ .

Calculation:

Simplify the expression by factoring the numerator and the denominator and diving out the common factors and then evaluate the limit.

  $\begin{array}{c}\underset{x\to -3}{\lim }\frac{x^2+7x+12}{x^2-9}=\underset{x\to -3}{\lim }\frac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x-3\right)}\\ =\underset{x\to -3}{\lim }\frac{x+4}{x-3}\\ =\frac{-3+4}{-3-3}\\ =-\frac{1}{6}\end{array}$

Thus, the given limit exists and is found as $\underset{x\to -3}{\lim }\frac{x^2+7x+12}{x^2-9}=-\frac{1}{6}$ .