Chapter 11.3, Problem 16E

Solution

a)

To explain: The reason behind the given statement.

After substitution, the limit is $-2$ and the denominator will be zero. Therefore, we can’t use substitution method.

Given information:

The given limit is $\underset{x\to -2}{{\displaystyle \lim }}\frac{\left|x^2-4\right|}{x+2}$ .

Calculation:

The given limit is $\underset{x\to -2}{{\displaystyle \lim }}\frac{\left|x^2-4\right|}{x+2}$ .

On substituting, the value $-2$ to the function the denominator will be zero.

b)

To find: The value of the limit $\underset{x\to -2}{{\displaystyle \lim }}\frac{\left|x^2-4\right|}{x+2}$ .

The limit $\underset{x\to -2}{{\displaystyle \lim }}\frac{\left|x^2-4\right|}{x+2}$ does not exist.

Given information:

The given limit is $\underset{x\to -2}{{\displaystyle \lim }}\frac{\left|x^2-4\right|}{x+2}$ .

Calculation:

Without taking modulus, we can simplify the expression as:

  $\frac{x^2-4}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)}=x-2$

Now,

  $f\left(x\right)=\left\{\begin{array}{l}\left|x-2\right|,x\ge 2\\ -\left|x-2\right|,x\prec -2\end{array}\right.$

As $x$ approaches $-2$ from the right,

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

As $x$ approaches $-2$ from the left,

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

The limit does not exist.