Standard Limits: Which Ones You Really Need and How to Spot Them

The standard limits — in Italian, limiti notevoli — that you really need for the Maturità are six: $\frac{\sin x}{x}$, $\frac{1-\cos x}{x^2}$, $\frac{\ln(1+x)}{x}$, $\frac{e^x-1}{x}$, $\frac{a^x-1}{x}$ and $\left(1+\frac{1}{x}\right)^x$. All the others follow from these with a substitution. The real difficulty is not learning them by heart, but recognising them when the question disguises them — and that is exactly what we watch our students struggle with every year between April and June.

This article goes deep on standard limits alone. If what you need is a general refresher on definitions, indeterminate forms and solution techniques, start from our complete guide to limits for the Maturità and then come back here.

Which standard limits do you really need for the Maturità?

Six formulas cover almost every case that appears in the questions and problems of the Maturità maths paper at a scientific high school. Here they are, each with the indeterminate form it starts from:

All the "cousins" you find on formula sheets — $\frac{\tan x}{x} \to 1$, $\frac{\arcsin x}{x} \to 1$, $\frac{(1+x)^k - 1}{x} \to k$ — come from these six with one algebraic step. Learning them as variants rather than as new formulas halves the memory load and cuts down on mix-ups under pressure.

Where do they turn up in the exam? Almost always in the single-answer questions ("evaluate the limit..."), and, in disguised form, in curve sketching: the behaviour of $f(x) = \frac{\sin 2x}{x}$ near the origin, for instance, is a standard limit dressed up as a graph.

How do you spot a disguised standard limit?

There is only one golden rule: the argument has to be the same above and below. The standard limit is not $\frac{\sin x}{x} \to 1$, it is $\frac{\sin(\text{something})}{\text{something}} \to 1$, provided that something tends to zero. The work in the exercise is almost always to rebuild that symmetry.

The method we teach in lessons has three steps:

1. Direct substitution. First of all, check that there really is an indeterminate form. If substituting gives you a number, you are done: no standard limits involved.
2. Find the tell. A $\sin$, a $\cos$, an $e^{(\cdot)} - 1$, a $\ln(1 + \cdot)$, a base tending to 1 with an exponent that blows up: each of these points to one particular standard limit (see the table below).
3. Balance the argument. Multiply and divide by whatever is missing, so that below (or in the exponent) you get exactly the argument of the function. This is where the marks are won or lost.

As an alternative to balancing, the substitution $t = \text{argument}$ always works: if you have $\sin 5x$, set $t = 5x$, rewrite everything in $t$ and the standard limit appears by itself. Slower, but error-proof: we recommend it to anyone who tends to lose pieces when working steps out in their head.

Decision table: from the indeterminate form to the standard limit

This is the table we get our students to build (and rewrite from memory) before the maths paper. Column one: what direct substitution gives you. Column two: the signal in the question. Column three: the tool to use.

A practical tip: while you work through exercises, before writing any step at all, note the indeterminate form in the margin in square brackets, like $\left[\tfrac{0}{0}\right]$. Students who skip this step tend to apply standard limits even where they are not needed — and there the mistake is not arithmetical but conceptual, which counts for more in an exam.

Three worked examples, with the numbers

Balancing the argument of the sine

The argument of the sine is $5x$, but below there is $3x$: we balance by multiplying and dividing by $5$.

The classic mistake here is answering $1$ "because it's sine over x". The standard limit equals $1$ only when the arguments match: here the answer is $\frac{5}{3} \approx 1.67$.

Two standard limits in the same problem

The tells: $e^{(\cdot)} - 1$ above, $\sin(\cdot)$ below. We balance both arguments:

Notice the mechanism: each standard limit "eats" its own argument and leaves nothing on the table but the ratio of the coefficients, $\frac{2}{3}$.

The $1^\infty$ form

We rewrite the base in the form $1 + \frac{1}{(\cdot)}$: here $\frac{3}{x} = \frac{1}{x/3}$, so the "right" exponent would be $\frac{x}{3}$. We adjust:

because $2x = \frac{x}{3} \cdot 6$. The answer is $e^6 \approx 403.4$: if you get $e^2$ or $e^3$, you have adjusted the base but not the exponent.

The mistakes we see most often in lessons

Marking Maturità mock papers in the weeks before the exam, the same four mistakes come back with striking regularity:

1. Using $\frac{\sin x}{x} \to 1$ with $x \to \infty$. The standard limit holds only for $x \to 0$. For $x \to +\infty$, $\sin x$ stays bounded between $-1$ and $1$ while the denominator blows up: the limit is $0$ by comparison. It is the conceptual mistake examiners forgive least, because it shows the formula has been memorised without being understood.
2. Forgetting the square with the cosine. $\frac{1-\cos x}{x^2} \to \frac{1}{2}$, but $\frac{1-\cos x}{x} \to 0$. They are different limits with different answers, and swapping the two denominators is probably the single most frequent slip we see on this topic.
3. Applying a standard limit where there is no indeterminate form. $\lim_{x \to \pi} \frac{\sin x}{x}$ is not a standard limit: direct substitution gives $\frac{0}{\pi} = 0$ straight away. Anyone who starts automatically with "sine over x is one" writes $1$ and loses the whole question.
4. Balancing the argument only halfway. In the limit $\frac{\sin 5x}{3x}$, multiplying only the denominator by $5$ (without compensating) changes the value of the limit. Every factor you put in has to go in as its reciprocal as well: multiply and divide, never one without the other.

Standard limits or L'Hôpital?

If your class syllabus includes L'Hôpital's rule, then technically many of these limits can also be solved by differentiating. There are two reasons to prefer the standard limits anyway. First, speed — $\frac{\sin 5x}{3x}$ closes in a single line with balancing. Second, and more subtle: using L'Hôpital on $\frac{\sin x}{x}$ is circular reasoning, because the derivative of $\sin x$ is proved from exactly that limit. An alert examiner can pull you up on it, and in an oral exam the question "why are you allowed to differentiate here?" is a classic.

The strategy we suggest: standard limits as first choice on $\frac{0}{0}$ forms with sines, exponentials and logarithms as $x \to 0$; L'Hôpital in reserve for the mixed cases where balancing gets unwieldy.

How to prepare between now and the exam

Standard limits stick through distributed practice, not through cramming: on our paths what works well is giving 10 minutes at the start of a lesson to 3-4 "disguised" limits picked at random, for weeks, until recognition becomes automatic. And they do not disappear after the Maturità: you will meet them again, unchanged, in first-year Analysis at university, so the time you invest now is worth double — and if you are already thinking about admission tests, take a look at our preparation paths.

If, on the other hand, you feel your gaps on limits and curve sketching are wider than this, a dedicated tutor can build you a plan targeted on the weeks left before the exam: tell us about your situation and book your first lesson.

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