Vertical, Horizontal and Oblique Asymptotes: How to Find Them With Limits

In curve sketching, asymptotes are worth more than the marks they carry: they are the skeleton of the graph. Get them wrong and the curve you draw at the end is wrong too, even if the derivative, the maxima and the points of inflection are perfect. In our fifth-year courses we see that almost nobody gets the calculation of the limit itself wrong: the mistake comes earlier, in deciding which limits to calculate, or later, in reading the result. This guide lines up the three types, the method for each one, four examples with the numbers, and the mistakes we correct most often.

What is an asymptote, and why does it matter in curve sketching?

An asymptote is a line that the graph gets closer and closer to as you move along the curve: either towards a point where the function "blows up", or towards infinity. There are three types — vertical $x = x_0$, horizontal $y = L$, oblique (or slant) $y = mx + q$ — and all three are found with a limit. That is why the chapter on limits comes before this one: if you need a refresher, our guide to limits for the Maturità starts from scratch.

In the calculus problem of the seconda prova, the Maturità maths paper, asymptotes are — after the domain and the sign — the first piece of the graph you can actually draw: you sketch the dashed lines and the curve has to slot in between them. That is why a missed or mistaken asymptote does not cost half a mark: it shifts the entire drawing. A useful detail for getting your bearings: a function can have infinitely many vertical asymptotes ($\tan x$ has one at every $x = \frac{\pi}{2} + k\pi$), but at most two between horizontal and oblique, one for each side.

How do you find a vertical asymptote?

The method has three steps. First: work out the domain, because the candidates are the excluded points and the finite edges — the zeros of the denominator, the point where the argument of a logarithm vanishes. Second: at each candidate $x_0$, calculate the right-hand limit and the left-hand limit, separately. Third: if at least one of the two is $\pm\infty$, the line $x = x_0$ is a vertical asymptote.

Take $f(x) = \dfrac{3x-1}{2x+5}$. Domain: $2x + 5 \neq 0$, that is $x \neq -\frac{5}{2}$. At $x_0 = -\frac{5}{2}$ the numerator is $3 \cdot \left(-\frac{5}{2}\right) - 1 = -\frac{17}{2}$, a number other than zero, while the denominator tends to $0$: the limit is infinite, and all that remains is to settle the sign.

For the sign of the denominator a test value is enough: to the right of $-2.5$, at $x = -2.4$ for instance, $2x + 5 = 0.2 > 0$; to the left, at $x = -2.6$, it is $-0.2 < 0$. Vertical asymptote $x = -\frac{5}{2}$, with the curve dropping to $-\infty$ as it comes in from the right and climbing to $+\infty$ as it comes in from the left.

The two separate limits do two jobs. They tell you which way the curve goes, without which the drawing is impossible. And they protect you from the cases where the asymptote exists on one side only: $\ln x$ has the asymptote $x = 0$ from the right alone, because to the left it is not even defined, and in a moment we shall see a function that blows up to the right of $0$ and quietly reaches the origin from the left.

Now the case that catches people out. $f(x) = \dfrac{x+2}{x^2-4}$. Domain: $x^2 - 4 \neq 0$, that is $x \neq 2$ and $x \neq -2$. Two candidates, but with different fates.

At $x = 2$ the numerator is $4$ and the denominator tends to $0$; $x^2 - 4$ is positive outside the interval $[-2, 2]$ and negative inside it, so:

Vertical asymptote $x = 2$. At $x = -2$, by contrast, numerator and denominator tend to $0$: the indeterminate form $\frac{0}{0}$, from which nothing follows. So we factorise:

A finite limit: no asymptote. At $x = -2$ the function has a hole, a removable discontinuity, at the point $\left(-2, -\frac{1}{4}\right)$: on the graph you mark it with an open circle, not with a dashed line.

How do you find a horizontal asymptote?

You calculate $\lim_{x \to +\infty} f(x)$ and $\lim_{x \to -\infty} f(x)$, one at a time. If the limit is a finite number $L$, the line $y = L$ is a horizontal asymptote on that side. For rational functions the result depends only on the degrees: numerator of lower degree, asymptote $y = 0$; equal degrees, $y$ equal to the ratio of the leading coefficients; numerator of higher degree, no horizontal asymptote (and you move on to look for the oblique one).

For $f(x) = \dfrac{3x-1}{2x+5}$ the degrees are equal. Taking $x$ out as a factor above and below:

Horizontal asymptote $y = \frac{3}{2}$, on both sides. For $f(x) = \dfrac{x+2}{x^2-4}$ the numerator has degree $1$ against $2$: the limit is $0$ and the asymptote is the $x$-axis, $y = 0$. You can see it from the simplified form $\frac{1}{x-2}$ as well.

The two sides need not agree. $\arctan x$ tends to $\frac{\pi}{2}$ as $x \to +\infty$ and to $-\frac{\pi}{2}$ as $x \to -\infty$: two different horizontal asymptotes. $e^x$ has the asymptote $y = 0$ only towards the left, while towards the right it climbs to infinity with no line to accompany it. Rational functions always behave the same way on the two sides; all the others — exponentials, arctangents, roots, absolute values — have to be checked one side at a time.

A detail that always comes as a surprise: the graph can cross a horizontal asymptote. $f(x) = \dfrac{x}{x^2+1}$ has the asymptote $y = 0$ and passes through the origin, so it crosses it at $x = 0$; $\dfrac{\sin x}{x}$ crosses it infinitely many times. A horizontal asymptote describes where the curve goes at infinity, not a wall somewhere finite. It is the vertical asymptote that cannot be crossed, because at $x_0$ the function is not defined.

When does an oblique asymptote exist, and how do you calculate $m$ and $q$?

You look for the oblique asymptote $y = mx + q$ only where there is no horizontal one, that is on a side where $f(x) \to \pm\infty$. On the same side, horizontal and oblique rule each other out: if you have found $y = \frac{3}{2}$ for $x \to +\infty$, looking for an oblique asymptote on the right as well is time wasted. For a rational function the condition is precise: the degree of the numerator equal to the degree of the denominator plus one. If the gap is two degrees or more, the function goes to infinity like a parabola or faster, and no line keeps up with it.

There are two formulas:

The asymptote exists if both limits are finite and $m \neq 0$. If $m$ comes out as $0$, or $q$ infinite, the function grows with no line to guide it: that is what happens with $\sqrt{x}$, with $\ln x$, with $x^2$.

Take $f(x) = \dfrac{2x^2+3x-1}{x-1}$. Degree $2$ above, $1$ below: a candidate. For completeness, the vertical asymptote is $x = 1$ (numerator $4$, hence $+\infty$ from the right and $-\infty$ from the left) and there is no horizontal one, because the limit at infinity is infinite.

Oblique asymptote $y = 2x + 5$, the same on both sides, as always for rational functions.

The step where marks get lost is the numerator of $q$: $-2x(x-1)$ is $-2x^2 + 2x$, and it is that $+2x$ that goes missing. Anyone who writes $2x^2 + 3x - 1 - 2x^2 - 2x = x - 1$ gets $q = 1$ and an asymptote $y = 2x + 1$: the right line shifted four units downwards, and a graph that no longer adds up.

There is a shorter and safer route, polynomial long division. Dividing $2x^2 + 3x - 1$ by $x - 1$ gives quotient $2x + 5$ and remainder $4$, that is

The quotient is the asymptote, and the remainder over the divisor is the part that tends to zero. A single line, with one extra piece of information into the bargain: $\frac{4}{x-1}$ is positive for $x > 1$ and negative for $x < 1$, so the curve sits above the line $y = 2x + 5$ to the right of the vertical asymptote and below it to the left. When you get to the drawing, that tells you which side of the dashed line to run the curve along.

A Maturità classic, from start to finish

$f(x) = x\,e^{1/x}$. It is not rational, so there are no shortcuts with the degrees: you follow the whole scheme.

Domain: $x \neq 0$. One candidate for the vertical asymptote, two directions for infinity.

Vertical, $x \to 0^+$: $\frac{1}{x} \to +\infty$, so $e^{1/x} \to +\infty$. The product is an indeterminate form $0 \cdot \infty$; setting $t = \frac{1}{x}$ it becomes $\lim_{t \to +\infty} \frac{e^t}{t} = +\infty$, because the exponential beats any power. Vertical asymptote $x = 0$, from the right.

Vertical, $x \to 0^-$: $\frac{1}{x} \to -\infty$, so $e^{1/x} \to 0$, and the product $x\,e^{1/x} \to 0$. From the left the curve reaches the origin with no asymptote at all. The same candidate line, opposite behaviour on the two sides: which is exactly why the limits are taken separately.

Horizontal: as $x \to \pm\infty$, $\frac{1}{x} \to 0$ and $e^{1/x} \to 1$, so $f(x)$ behaves like $x$ and tends to $\pm\infty$. No horizontal asymptote, so we move on to the oblique one.

The last step is the standard limit $\frac{e^t - 1}{t} \to 1$, which you will find explained in our guide to standard limits for the Maturità. Oblique asymptote $y = x + 1$ on both sides.

To sum up: vertical asymptote $x = 0$ from the right only, no horizontal asymptote, oblique $y = x + 1$. With these three pieces of information and the sign of the function, the graph is already recognisable before you have even touched the derivative.

The mistakes we see most often in lessons

Marking fifth-year class tests and Maturità mock papers, six mistakes come back with more regularity than any others:

1. The limit taken from one side only. $\lim_{x \to x_0}$ gets calculated without distinguishing $x_0^-$ from $x_0^+$, and written as "$= \infty$" with no sign. Then you do not know which way to draw the curve, and in cases like $x\,e^{1/x}$ an asymptote on the left gets invented that does not exist.
2. An excluded point mistaken for an asymptote. $\frac{x+2}{x^2-4}$ declared to have two vertical asymptotes. At $-2$ there is a hole. The check is to substitute $x_0$ into the numerator as well: if you get zero, factorise before concluding.
3. A vertical asymptote written as $y = \ldots$ A vertical asymptote is a vertical line, and its equation is $x = x_0$. "Vertical asymptote $y = 2$" is an error of form that some examiners penalise, and one that shows the line was never actually pictured.
4. The oblique asymptote hunted for where the horizontal one is, or forgotten where it is needed. If $\lim_{x \to +\infty} f(x)$ is finite, there is no oblique asymptote on the right: $m$ and $q$ should not even be calculated. If instead the limit is infinite and, for rational functions, the numerator exceeds the denominator by exactly one degree, the oblique asymptote does have to be looked for. Skipping it leaves the graph without guidance precisely where it needs it.
5. The sign in the numerator of $q$. Calculating $f(x) - mx$ means subtracting a whole polynomial: the minus has to be distributed over every term. It is the mistake that turns $y = 2x + 5$ into $y = 2x + 1$. If in doubt, do the polynomial division and compare the two results.
6. Only one infinity checked. For $e^x$, $\arctan x$, and for functions with roots or absolute values, the two sides can give different results. The limit has to be calculated for $+\infty$ and for $-\infty$, always, even when the exercise looks symmetric.

From asymptotes to the complete graph

Domain, sign and asymptotes are the first draft of the graph: before you differentiate anything you already know which regions of the plane the curve lives in and which lines it stretches towards. That is why, in curve sketching, it pays to draw the dashed lines straight away, before moving on to the derivative — if the sign of $f'$ then sends the curve against a vertical asymptote from the wrong side, the mistake shows up on its own. The same limits at infinity reappear, unchanged, in the calculus questions of science admission tests and in your first university analysis exam: if you are already looking beyond the Maturità, take a look at our preparation paths.

If instead curve sketching feels like a list of steps with no logic behind it, the problem usually lies further back, in limits or in the algebra of fractions, and it has to be rebuilt from there: tell us about your situation and book your first lesson with a tutor who picks up from the point where the thread broke.

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