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 \(f(x)\).