Some curves run off to infinity in a predictable way: they snuggle up to a straight line, getting ever closer without (necessarily) touching it. These lines are called asymptotes, and finding them is one of the first steps in sketching a graph. This post explains what an asymptote is, how to find vertical, horizontal and oblique ones, and the traps to avoid.
1. What is an asymptote?
Definition
A straight line \( Ax + By + C = 0 \) is called an asymptote of the curve \( y = f(x) \) if the distance from a point moving along the curve to this line tends to zero as \(x\) or \(y\) tends to infinity.
Two things must happen together: the point runs off to infinity (in \(x\), in \(y\), or both), and its distance to the line shrinks to zero. A curve that just stays close to a line on a bounded stretch has no asymptote there; the closeness has to last "forever".
For the graph of a function there are two kinds of asymptotes:
vertical \(x = a\): the point runs off to infinity in \(y\), while \(x\) approaches a finite value \(a\);
oblique \(y = kx + n\) (horizontal when \(k = 0\)): the point runs off to infinity in \(x\).
(The graph of a function cannot have any other kind: a line \(x = a\) is the only line you can approach while \(x\) stays bounded.)
2. Vertical asymptotes
Vertical asymptote
If there is a number \(a\) such that
\[ \lim_{x\to a\pm0} f(x) = \pm\infty, \]
then the line \(x = a\) is a vertical asymptote.
Here the distance from the point \( (x, f(x)) \) to the line \(x = a\) is simply \( |x - a| \), which tends to \(0\), while \(f(x)\) runs to infinity. It is enough that one of the one-sided limits (\(x \to a - 0\) or \(x \to a + 0\)) is infinite.
Where to look: a continuous function cannot blow up at a point where it is defined and continuous. So the candidates for \(a\) are the boundary points of the domain and the points of discontinuity:
zeros of a denominator, as for \( \frac{1}{x - 2} \) at \(x = 2\);
endpoints of the domain of a logarithm, as for \( \ln x \) at \(x = 0\);
points like \( \frac\pi2 + k\pi \) for \( \operatorname{tg}x \).
Example 1
\( f(x) = \frac{x}{x^2 - 1} \): at \(x = \pm1\) the denominator vanishes and the numerator doesn't, so \(f \to \pm\infty\). Vertical asymptotes \(x = 1\) and \(x = -1\).
\( f(x) = \ln x \): \( \lim_{x\to0^+}\ln x = -\infty \). Vertical asymptote \(x = 0\) (approached only from the right).
\( f(x) = \frac{\sin x}{x} \): at \(x = 0\) the denominator vanishes, but \( \frac{\sin x}{x} \to 1 \). No asymptote: a zero denominator is only a candidate, and the limit decides.
3. Oblique and horizontal asymptotes
Oblique asymptote
If the limits
\[ k = \lim_{x\to\pm\infty}\frac{f(x)}{x} \qquad \text{and} \qquad n = \lim_{x\to\pm\infty}\big[f(x) - kx\big] \]
exist (and are finite), then the line \( y = kx + n \) is an oblique asymptote. For \( k = 0 \) it is horizontal: \( y = n \), with \( n = \lim_{x\to\pm\infty}f(x) \).
Where the formulas come from. The line \( y = kx + n \) is an asymptote as \(x \to \infty\) exactly when the vertical gap between curve and line vanishes:
(The true distance to the line is this vertical gap times \( \cos\alpha \), where \(\alpha\) is the angle of the line; since \(\cos\alpha\) is a fixed nonzero number, the two tend to zero together.)
Dividing by \(x\): \( \frac{f(x)}{x} - k - \frac nx \to 0 \). Since \( \frac nx \to 0 \), we must have \( k = \lim\frac{f(x)}{x} \). So \(k\) is the slope the curve settles into far away.
Once \(k\) is known, the condition itself says \( n = \lim\big[f(x) - kx\big] \): the intercept.
Conversely, if both limits exist, then \( f(x) - kx - n \to 0 \), so the line really is an asymptote. Both limits must exist; if either fails (or is infinite), there is no oblique asymptote on that side.
4. A complete example
Example 2 · \( f(x) = \dfrac{x^2 + 1}{x - 1} \)
Vertical. The denominator vanishes at \(x = 1\), and the numerator equals \(2\) there:
So \( y = x + 1 \) is an oblique asymptote, on both sides.
Which side of the asymptote? The gap is \( f(x) - (x + 1) = \frac{2}{x - 1} \): positive for \(x \to +\infty\) (curve above the line), negative for \(x \to -\infty\) (below). Checking this sign is a good habit when sketching.
Figure 1. \( y = \frac{x^2 + 1}{x - 1} \) (blue) with its vertical asymptote \(x = 1\) and its oblique asymptote \(y = x + 1\) (red, dashed). The right branch approaches the oblique asymptote from above, the left branch from below.
5. Plus and minus infinity can differ
Notice the \(\pm\) in the formulas: the limits must be computed separately for \( x \to +\infty \) and \( x \to -\infty \). A curve can have different asymptotes on the two sides, or an asymptote on one side only.
Example 3 · Two horizontal asymptotes
\( f(x) = \operatorname{arctg}x \): \( \lim_{x\to+\infty} = \frac\pi2 \), \( \lim_{x\to-\infty} = -\frac\pi2 \). Horizontal asymptotes \( y = \frac\pi2 \) on the right and \( y = -\frac\pi2 \) on the left.
Example 4 · Two oblique asymptotes
\( f(x) = \sqrt{x^2 + 1} \). For \(x \to +\infty\): \( k = \lim\frac{\sqrt{x^2 + 1}}{x} = 1 \), and
\[ n = \lim_{x\to+\infty}\big(\sqrt{x^2 + 1} - x\big) = \lim_{x\to+\infty}\frac{1}{\sqrt{x^2 + 1} + x} = 0. \]
For \(x \to -\infty\): \( \frac{\sqrt{x^2 + 1}}{x} = \frac{|x|\sqrt{1 + 1/x^2}}{x} \to -1 \), and similarly \(n = 0\). So \( y = x \) on the right and \( y = -x \) on the left: the curve is the upper branch of the hyperbola \( y^2 - x^2 = 1 \).
Note: \( \sqrt{x^2} = |x| \), which equals \(-x\) for negative \(x\). Forgetting this is the most common error in such problems.
Example 5 · An asymptote on one side only
\( f(x) = e^x \): \( \lim_{x\to-\infty}e^x = 0 \), so \(y = 0\) is a horizontal asymptote on the left. On the right, \( k = \lim\frac{e^x}{x} = \infty \): no asymptote. The exponential grows faster than any line.
Example 6 · Everything at once
\( f(x) = x\,e^{1/x} \).
Vertical: as \(x \to 0^+\), \( x\,e^{1/x} = \frac{e^{t}}{t} \to +\infty \) (\(t = \frac1x \to +\infty\)). As \(x \to 0^-\), \( e^{1/x} \to 0 \) and \(f \to 0\). So \(x = 0\) is a vertical asymptote, approached only from the right.
Oblique: \( k = \lim e^{1/x} = 1 \) and, with \(t = \frac1x \to 0\), \( n = \lim x\big(e^{1/x} - 1\big) = \lim\frac{e^t - 1}{t} = 1 \) (L'Hôpital). So \(y = x + 1\), on both sides.
6. Rational functions: asymptotes by division
For a rational function \( f(x) = \frac{P(x)}{Q(x)} \), there is a faster method than computing limits: divide the polynomials.
If \(\deg P < \deg Q\): horizontal asymptote \( y = 0 \).
If \(\deg P = \deg Q\): horizontal asymptote \( y = \frac{\text{leading coefficient of }P}{\text{leading coefficient of }Q} \).
If \(\deg P = \deg Q + 1\): divide, \( f(x) = kx + n + \frac{R(x)}{Q(x)} \) with \( \deg R < \deg Q \). The quotient \( y = kx + n \) is the oblique asymptote, because the remainder term tends to \(0\).
If \(\deg P \ge \deg Q + 2\): no horizontal or oblique asymptote (the curve behaves like a parabola or steeper).
Vertical asymptotes are at the zeros of \(Q\) that are not cancelled by zeros of \(P\).
Example 7
Example 2 again: \( \frac{x^2 + 1}{x - 1} = x + 1 + \frac{2}{x - 1} \). The asymptote \(y = x + 1\) and even the sign of the gap \( \frac{2}{x - 1} \) can be read off directly.
7. Pitfalls
\(k\) may exist while \(n\) does not
For \( f(x) = x + \sqrt x \): \( k = \lim\frac{x + \sqrt x}{x} = 1 \), but \( n = \lim\sqrt x = \infty \). No asymptote. The curve's slope tends to \(1\), yet it drifts ever further from every line of slope \(1\). Similarly \( \ln x \) has \(k = 0\) but \(n = \infty\): it flattens out without approaching any horizontal line. Always compute both limits.
A curve may cross its asymptote
It is a myth that a curve can never touch its asymptote. \( f(x) = \frac{\sin x}{x} \) has the horizontal asymptote \(y = 0\) and crosses it infinitely many times, at \(x = k\pi\). What matters is only the behaviour far away: the oscillations shrink to zero. (Vertical asymptotes of a function, on the other hand, are never crossed, because the function is not defined at \(x = a\) or jumps there.)
Undefined does not mean asymptote
\( f(x) = \frac{x^2 - 4}{x - 2} \) is undefined at \(x = 2\), but \( f(x) = x + 2 \) for \(x \ne 2\), with limit \(4\). It's a hole in the graph, not a vertical asymptote. The zero of the denominator was cancelled by a zero of the numerator.
8. Summary
Type
Condition
Where to look
Vertical \(x = a\)
\( \lim_{x\to a\pm0}f(x) = \pm\infty \) (one side is enough)
zeros of denominators, boundary of the domain
Oblique \(y = kx + n\)
\( k = \lim\frac{f(x)}{x} \), \( n = \lim[f(x) - kx] \), both finite
separately at \(+\infty\) and \(-\infty\)
Horizontal \(y = n\)
\(k = 0\), \( n = \lim f(x) \)
special case of oblique
Checklist:
Find the domain. At each boundary point and point of discontinuity, compute the one-sided limits: vertical asymptotes.
For \(x \to +\infty\): compute \(k\), then \(n\). Repeat for \(x \to -\infty\).
For rational functions, divide the polynomials instead.
Optional but useful: the sign of \( f(x) - (kx + n) \) tells on which side of the asymptote the curve lies.
9. Practice problems
Try each one before opening the solution. More solved problems from this topic are in Differential Calculus.
Problem 1: Find the asymptotes of \( y = \dfrac{2x^2 - 3}{x + 1} \).
Problem 5: Find the asymptotes of \( y = \sqrt{x^2 - 2x} \).
For \(x \to +\infty\): \( k = 1 \), \( n = \lim\big(\sqrt{x^2 - 2x} - x\big) = \lim\frac{-2x}{\sqrt{x^2 - 2x} + x} = -1 \): \(y = x - 1\). For \(x \to -\infty\): \( k = -1 \), \( n = \lim\big(\sqrt{x^2 - 2x} + x\big) = 1 \): \(y = -x + 1\).
Problem 6: Does \( y = x\sin\dfrac1x \) have asymptotes?
At \(x = 0\): \( |x\sin\frac1x| \le |x| \to 0 \), no vertical asymptote. As \(x \to \pm\infty\): \( x\sin\frac1x = \frac{\sin t}{t} \to 1 \) (\(t = \frac1x\)). Horizontal \(y = 1\) on both sides.
Problem 7: Find the asymptotes of \( y = \dfrac{x - 1}{x + 1} \) and determine on which side of the horizontal asymptote the curve lies.
\( y = 1 - \frac{2}{x + 1} \). Vertical \(x = -1\), horizontal \(y = 1\). For \(x \to +\infty\) the gap \( -\frac{2}{x + 1} < 0 \): below the asymptote; for \(x \to -\infty\): above it.
Problem 8: Why doesn't \( y = x^2 \) have an oblique asymptote?
\( k = \lim\frac{x^2}{x} = \pm\infty \). The parabola gets steeper and steeper; no fixed slope can follow it.
Final thoughts
Asymptotes describe a curve's behaviour at infinity: where it shoots up next to a vertical line, and which straight line it eventually follows far to the left and right. Two limits, \(k\) for the slope and \(n\) for the intercept, do all the work, computed separately on each side. Together with monotonicity, extrema and concavity from the previous posts, asymptotes give you everything needed to sketch the graph of a function from its formula.