27 December 2021

Statement

Find the vertical and horizontal asymptotes of \(f(x) = \frac{x^3}{x^2 + 3x - 10}\)

Solution

By looking at the higher degree polynomial in the numerator we can take a guess that \(f(x)\) diverges when \(x \to \infty\)

When \(x \to +\infty\)

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

\(= \lim\limits_{x \to +\infty} \frac{x^3}{x^2 + 3x - 10}\)

\(= \lim\limits_{x \to +\infty} \frac{x^3}{x^2 (1 + \frac{3}{x} - \frac{10}{x^2})}\)

\(= \lim\limits_{x \to +\infty} \frac{x}{1 + \frac{3}{x} - \frac{10}{x^2}}\)

\(= \frac{\lim\limits_{x \to +\infty} x}{ \lim\limits_{x \to +\infty} 1 + \frac{3}{x} - \frac{10}{x^2}}\)

\(= \frac{\infty}{1}\)

\(= + \infty\)


When \(x \to -\infty\)

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

\(= \lim\limits_{x \to -\infty} \frac{x^3}{x^2 + 3x - 10}\)

\(= \lim\limits_{x \to -\infty} \frac{x^3}{x^2 (1 + \frac{3}{x} - \frac{10}{x^2})}\)

\(= \lim\limits_{x \to -\infty} \frac{x}{1 + \frac{3}{x} - \frac{10}{x^2}}\)

\(= \frac{\lim\limits_{x \to -\infty} x}{ \lim\limits_{x \to -\infty} 1 + \frac{3}{x} - \frac{10}{x^2}}\)

\(= \frac{\infty}{1}\)

\(= + \infty\)


This confirms our initial guess that \(f(x)\) diverges when \(x \to \pm \infty\), hence not having any horizontal asymptote. To find the vertical asymptotes we need to find values of \(x\) such that the denominator \(x^2 + 3x - 10 = 0\).

\(x^2 + 3x - 10 = 0\)

\(\Leftrightarrow x = \frac{-3 \pm \sqrt{9 - 4 \times 1 \times (-10)}}{2}\)

\(\Leftrightarrow x = \frac{-3 \pm \sqrt{49}}{2}\)

\(\Leftrightarrow x = \frac{-3 \pm 7}{2}\)

\(\Leftrightarrow x = -5 \lor x = 2\)


This indicates that \(f(x)\) has vertical asymptotes at \(x = -5\) and \(x = 2\). We can confirm this by the following calculations.

\(\lim\limits_{x \to 2^{-}} \frac{x^3}{x^2 + 3x -10} = \frac{8}{4^{-} + 6^{-} - 10} = \frac{8}{0^{-}} = -\infty\)

\(\lim\limits_{x \to 2^{+}} \frac{x^3}{x^2 + 3x -10} = \frac{8}{4^{+} + 6^{+} - 10} = \frac{8}{0^{+}} = +\infty\)

\(\lim\limits_{x \to -5^{-}} \frac{x^3}{x^2 + 3x -10} = \frac{-125}{25^{+} - 15^{-} - 10} = \frac{-125}{0^{+}} = -\infty\)

\(\lim\limits_{x \to -5^{-}} \frac{x^3}{x^2 + 3x -10} = \frac{-125}{25^{-} - 15^{+} - 10} = \frac{-125}{0^{-}} = +\infty\)


Which means \(f(x)\) has vertical asymptotes at \(x = -5\) and \(x = 2\).