Chapter 1, Problem 27RE

(a)

To determine

To find: The vertical asymptote of the graph of $y=f(x)$

Answer to Problem 27RE

The vertical asymptote of the graph of $y=f(x)$ is $x=-2$

Explanation of Solution

Given information: The function is $f(x)=\frac{x+3}{x+2}$

Calculation:

The given function is:

  $f(x)=\frac{x+3}{x+2}$

Given that $f$ is not defined at $x=-2$ due to division by zero and $x+2$ cannot be factored out of the function formula then $x=-2$ is a vertical asymptote of the graph of $y=f(x)$ .

(b)

To determine

To describe: The behavior of $f(x)$ to the left and right of any vertical asymptote.

Answer to Problem 27RE

  $\underset{x\to -2^{+}}{\lim }f(x)=\infty$ and

  $\underset{x\to -2^{-}}{\lim }f(x)=-\infty$

Explanation of Solution

Given information: The function is $f(x)=\frac{x+3}{x+2}$

Calculation:

The given function is:

  $f(x)=\frac{x+3}{x+2}$

To examine the behavior of $f$ on both sides of the asymptote .When $x$ approaches $-2$ from the positive side, it will be taken on values such as

  $\begin{array}{c}x=-1.9\\ x=-1.99\\ x=-1.999,\dots .\end{array}$

For such values the signs of the numerator and denominator will be both positive and so $f(x)$ will become a larger and larger positive number , Hence

  $\underset{x\to -2^{+}}{\lim }f(x)=\infty$

When $x$ approaches $-2$ from the negative side, it will be taken on values such as

  $\begin{array}{c}x=-2.1\\ x=-2.01\\ x=-2.001,\dots \end{array}$

For such values the signs of the numerator will be positive and denominator will be negative and so $f(x)$ will become a larger and larger negative number , Hence

  $\underset{x\to -2^{-}}{\lim }f(x)=-\infty$