Asymptotes
Summarize
Horizontal asymptotes
An asymptote of a function is a line that its graph approaches arbitrarily closely, either as the variable tends to infinity or near a point of discontinuity. The definition relies entirely on the notion of limit.
Let $y = f(x)$ be a real-valued function defined on an interval of the form $[a, +\infty)$, $(-\infty, b]$, or on $\mathbb{R}$. The line with equation $y = L$ is a horizontal asymptote of $f$ if:
$$\lim_{x \to +\infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L$$
Consider the function:
$$y = \frac{x + 1}{x}$$
By computing the limit as $x \to \pm\infty$, we obtain:
$$\lim_{x \to \pm\infty} \frac{x + 1}{x} = \lim_{x \to \pm\infty} \left( 1 + \frac{1}{x} \right) = 1$$
The function therefore admits the horizontal asymptote $y = 1$, which both branches of the graph approach as $x \to +\infty$ and $x \to -\infty$.
Vertical asymptotes
Let $y = f(x)$ be a real-valued function defined on $[a, b] \setminus \\{ x_0 \\}$, where $x_0 \in [a, b]$. The line with equation $x = x_0$ is a vertical asymptote of $f$ if:
$$\lim_{x \to x_0^-} f(x) = \pm \infty \quad \text{or} \quad \lim_{x \to x_0^+} f(x) = \pm \infty$$
A standard example is the function:
$$y = \frac{1}{x - 1}$$
This rational expression is undefined at $x = 1$, since the denominator becomes zero. To analyse the behaviour of $f(x)$ near $x = 1$, we compute the one-sided limits:
$$\lim_{x \to 1^-} \frac{1}{x - 1} = -\infty \qquad \lim_{x \to 1^+} \frac{1}{x - 1} = +\infty$$
The function diverges to $-\infty$ when approaching $1$ from the left, and to $+\infty$ when approaching from the right. The line $x = 1$ is therefore a vertical asymptote of $f$.
A rational function often has vertical asymptotes where its denominator vanishes and the function is undefined. Such points are non-removable discontinuities, where the function fails to be continuous.
Oblique asymptotes
Let $y = f(x)$ be a real-valued function defined on a half-line of the form $(-\infty, a]$ or $[a, +\infty)$. The line with equation $y = p x + q$ is an oblique asymptote of $f$ if:
$$ \begin{align} \lim_{x \to -\infty} \big[ f(x) - (p x + q) \big] &= 0 \\[6pt] \lim_{x \to +\infty} \big[ f(x) - (p x + q) \big] &= 0 \end{align} $$
The graph of $f$ therefore approaches the line $y = p x + q$ for large values of $|x|$.
The equation of the oblique asymptote $y = p x + q$ associated with a function $f(x)$ is determined by computing two specific limits. The slope $p$ is found by evaluating:
$$p = \lim_{x \to \pm\infty} \frac{f(x)}{x}$$
Once the slope is known, the vertical offset $q$ is obtained from:
$$q = \lim_{x \to \pm\infty} \big[ f(x) - p x \big]$$
If both limits exist and are finite, then the line $y = p x + q$ is the oblique asymptote of the function.
Consider the function:
$$f(x) = \frac{x^2 + 1}{x}$$
To determine whether this function admits an oblique asymptote as $x \to \pm\infty$, we begin with the limit of $f(x)/x$:
$$\frac{f(x)}{x} = \frac{x^2 + 1}{x^2} = 1 + \frac{1}{x^2}$$
As $x \to \pm\infty$, the term $\dfrac{1}{x^2}$ tends to zero, so:
$$\lim_{x \to \pm\infty} \frac{f(x)}{x} = 1$$
The slope of the asymptote is therefore $p = 1$. Next, we compute the limit of the difference $f(x) - p x$:
$$f(x) - x = \frac{x^2 + 1}{x} - x = \frac{x^2 + 1 - x^2}{x} = \frac{1}{x}$$
Since $\frac{1}{x} \to 0$ as $x \to \pm\infty$, we obtain:
$$\lim_{x \to \pm\infty} [f(x) - x] = 0$$
The function therefore admits an oblique asymptote with equation:
$$y = x$$
In this case the oblique asymptote passes through the origin, which gives $q = 0$, a degenerate configuration. In general, the vertical offset $q$ need not be zero.
Example 1
Here the vertical offset $q$ is nonzero, unlike the previous case. Take the function:
$$f(x) = \frac{2 x^2 - x + 3}{2 x}$$
To determine the oblique asymptote, we first compute the slope $p$:
$$ \begin{align} p &= \lim_{x \to \pm\infty} \frac{f(x)}{x} \\[6pt] &= \lim_{x \to \pm\infty} \frac{2 x^2 - x + 3}{2 x^2} \\[6pt] &= 1 \end{align} $$
Next, we evaluate the vertical offset $q$:
$$ \begin{align} q &= \lim_{x \to \pm\infty} \big( f(x) - x \big) \\[6pt] &= \lim_{x \to \pm\infty} \frac{2 x^2 - x + 3 - 2 x^2}{2 x} \\[6pt] &= \lim_{x \to \pm\infty} \frac{- x + 3}{2 x} \\[6pt] &= -\frac{1}{2} \end{align} $$
The equation of the oblique asymptote is therefore:
$$y = x - \frac{1}{2}$$
Properties of asymptotes
Several properties follow from these definitions:
- Not every function admits asymptotes.
- For horizontal asymptotes, several configurations are possible: a function may have none, it may approach the same horizontal line as $x \to +\infty$ and $x \to -\infty$, or it may approach two distinct horizontal lines in the two directions.
- Different types of asymptotes can coexist. A single function may simultaneously exhibit horizontal, vertical, and oblique asymptotes, depending on its behaviour near discontinuities and at infinity.
- Vertical asymptotes are not confined to rational functions: the logarithm $\ln x$ has one at $x = 0$, and $\tan x$ has infinitely many.
- A graph may cross its horizontal asymptote rather than only approach it, as $\frac{\sin x}{x}$ does infinitely often near $y = 0$.
- For a rational function, an oblique asymptote occurs when the degree of the numerator exceeds that of the denominator by exactly one.
The asymptote condition $f(x) - (p x + q) \to 0$ can be written $f(x) = p x + q + o(1)$ in little-o notation. Landau symbols more generally compare the growth rates of functions.