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 $$x = 2$$.