27 December 2021

Statement

Find the vertical and horizontal asymptotes of $$f(x) = \frac{x}{x + 4}$$

Solution

Let's start by checking if $$f$$ has any horizontal asymptotes.

When $$x \to +\infty$$

$$\lim\limits_{x \to +\infty} f(x)$$

$$= \lim\limits_{x \to +\infty} \frac{x}{x + 4}$$

$$= \lim\limits_{x \to +\infty} \frac{x}{x (1 + \frac{4}{x})}$$

$$= \lim\limits_{x \to +\infty} \frac{1}{1 + \frac{4}{x}}$$

$$= \frac{1}{1 + 0}$$

$$= 1$$

When $$x \to -\infty$$

$$\lim\limits_{x \to -\infty} f(x)$$

$$= \lim\limits_{x \to -\infty} \frac{x}{x + 4}$$

$$= \lim\limits_{x \to -\infty} \frac{x}{x (1 + \frac{4}{x})}$$

$$= \lim\limits_{x \to -\infty} \frac{1}{1 + \frac{4}{x}}$$

$$= \frac{1}{1 + 0}$$

$$= 1$$

Therefore $$f(x)$$ has a horizontal asymptote when $$y = 1$$. To find the vertical asymptotes let's look for values of $$x$$ where $$f(x)$$ is not defined. When $$x = -4$$ the denominator of $$f(x)$$ is zero, which indicates that when $$x \to -4$$ $$f(x)$$ approaches infinity. Let's confirm this by finding the limits from both sides.

From the right side

$$\lim\limits_{x \to -4^{+}} \frac{x}{x + 4}$$

$$= \lim\limits_{x \to -4^{+}} \frac{-4}{-4^{+} + 4}$$

$$= \lim\limits_{x \to -4^{+}} \frac{-4}{0^{+}}$$

$$= -\infty$$

From the left side

$$\lim\limits_{x \to -4^{-}} \frac{x}{x + 4}$$

$$= \lim\limits_{x \to -4^{-}} \frac{-4}{-4^{-} + 4}$$

$$= \lim\limits_{x \to -4^{-}} \frac{-4}{0^{-}}$$

$$= +\infty$$

As we can see $$f(x)$$ has a vertical asymptote at $$x = -4$$. Since $$f(x)$$ is continuous at any other value of $$x$$ we can be certain this is the only vertical asymptote of $$f(x)$$.