How to Find Horizontal and Vertical Asymptotes Using Limits

Both kinds of asymptote are defined by limits. The method is to compute the right limits, not to memorise shortcuts, although the shortcuts for rational functions follow from those limits.

Definitions

Horizontal asymptote. The line $y = L$ is a horizontal asymptote of $f$ if

$$\lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L.$$

The two directions can give different values, so a function can have two different horizontal asymptotes.

Vertical asymptote. The line $x = a$ is a vertical asymptote of $f$ if at least one of

$$\lim_{x \to a^-} f(x), \qquad \lim_{x \to a^+} f(x)$$

is $+\infty$ or $-\infty$.

Method

1. Vertical: find the points where $f$ is undefined or “blows up”, usually zeros of a denominator or of the argument of a logarithm. At each candidate $a$, compute the one-sided limits.
2. Horizontal: compute $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$. For rational functions, divide numerator and denominator by the highest power of $x$ in the denominator.

Worked example 1: a rational function

$$f(x) = \frac{2x^2 + 3}{x^2 - 4}$$

Horizontal. Divide top and bottom by $x^2$:

$$\lim_{x \to \pm\infty} \frac{2 + 3/x^2}{1 - 4/x^2} = \frac{2 + 0}{1 - 0} = 2.$$

So $y = 2$ is a horizontal asymptote in both directions.

Vertical. The denominator $x^2 - 4 = (x-2)(x+2)$ is zero at $x = \pm 2$, and the numerator there is $2(4) + 3 = 11 \neq 0$. Check the signs at $x = 2$:

• $x \to 2^+$: the denominator is a small positive number, so $f(x) \to +\infty$.
• $x \to 2^-$: the denominator is a small negative number, so $f(x) \to -\infty$.

So $x = 2$ is a vertical asymptote, and by symmetry ($f$ is even) so is $x = -2$.

Worked example 2: different limits at ±∞

$$g(x) = \frac{3x}{\sqrt{x^2 + 1}}$$

For $x > 0$, $\sqrt{x^2 + 1} = x\sqrt{1 + 1/x^2}$, so $g(x) = \dfrac{3}{\sqrt{1 + 1/x^2}} \to 3$.

For $x < 0$, $\sqrt{x^2} = |x| = -x$, so $\sqrt{x^2+1} = -x\sqrt{1 + 1/x^2}$ and $g(x) \to -3$.

So $g$ has two horizontal asymptotes: $y = 3$ as $x \to \infty$ and $y = -3$ as $x \to -\infty$. It has no vertical asymptote, because the denominator is never zero.

Worked example 3: a hole, not an asymptote

$$h(x) = \frac{x^2 - 1}{x - 1}$$

The denominator vanishes at $x = 1$, but $h(x) = x + 1$ for $x \ne 1$, so $\lim_{x\to1} h(x) = 2$, which is finite. The graph has a removable discontinuity (a hole) at $(1, 2)$, not a vertical asymptote. Always cancel common factors first.

Shortcuts for rational functions

For $f = P/Q$ with $\deg P = m$ and $\deg Q = n$:

Degrees	Horizontal asymptote
$m < n$	$y = 0$
$m = n$	$y = \dfrac{\text{leading coefficient of } P}{\text{leading coefficient of } Q}$
$m > n$	none. If $m = n + 1$, there is a slant asymptote, found by polynomial division.

These are just the results of the limit calculation above.

Common mistakes

• Only checking $x \to \infty$. Square roots and exponentials often behave differently as $x \to -\infty$.
• Declaring every zero of the denominator an asymptote without checking for cancellation.
• Thinking a graph can’t cross a horizontal asymptote. It can: $\dfrac{\sin x}{x}$ crosses $y = 0$ infinitely often. A graph can never cross a vertical asymptote, because $f$ is undefined there.