Chapter 11.3, Problem 15E

Solution

a)

To explain: Why you cannot use substitution to find the limit $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}$ .

When one substitutes -2 for $x$ then the denominator becomes zero, so one cannot use substitution to find the limit $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}$ .

Given information: The given limit is $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}$ .

Calculation:

Substitute $-2$ for $x$ in the given limit then the denominator becomes zero as shown below:

  $\begin{array}{c}x+2=-2+2\\ =0\end{array}$

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

b)

To find: The limit $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}$ algebraically if it exists.

The limit $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}$ exists and is found by factoring numerator as $-4$ .

Given information: The given limit is $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}$ .

Calculation:

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

  $\begin{array}{c}\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}=\underset{x\to -2}{\lim }\frac{\left(x+2\right)\left(x-2\right)}{x+2}\\ =\underset{x\to -2}{\lim }\left(x-2\right)\\ =-2-2\\ =-4\end{array}$

Thus, the given limit exists and is found as $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}=-4$ .