Limits and indeterminate forms: how to solve them

In Italian technical institutes, whether the economics or the technology track, maths has three hours a week and the Year 4 syllabus reaches limits between November and December. The test usually falls in December, six to nine weeks from now. Until then your work is something else: studying functions and domains. When the teacher writes “calculate the limit” on the board, the real calculation is almost always a case of an indeterminate form, and that is where the grade is decided.

This post is about technique: given a limit, how to recognise the form and which move is needed. The definitions and ideas are elsewhere (limits explained well), asymptotes and continuity too (limits, continuity and asymptotes and limits and asymptotes in function study). For the picture of the year there is the Year 4 formula sheet.

The first step: substitute

Every limit begins by substituting the value that $x$ tends to. It is not a formality: in most cases the limit comes out at once, and recognising it saves you half a sheet. Substitutions give results that are not indeterminate forms:

where $k$ is a non-zero number. When the denominator tends to zero with $k\ne 0$ you need the one-sided limits, to establish whether the result is $+\infty$ or $-\infty$, but the form is not indeterminate.

There are seven indeterminate forms. In the Year 4 syllabus you mostly meet four:

The other three ($1^{\infty }$, $0^0$, ${\infty }^0$) appear with powers and are usually solved with a standard limit, or do not reach the technical-school test at all.

The name “indeterminate” does not mean the limit does not exist. It means that from substitution alone you cannot know, because it depends on how the two parts tend to their value. That is why you rewrite the expression until the obstacle is removed.

The scheme: what form it is, what technique is needed

The line that counts is the first column: every time you get stuck, go back and ask yourself what form it is, then see whether there are roots or transcendental functions. Usually the technique is just one, and the work lies in recognising it.

$\frac{\infty }{\infty }$ with rational functions

When $x\to \pm \infty$, the highest-degree term of numerator and denominator rules. You factor out $x$ to the highest power and the rest tends to zero. Comparing the degrees gives the result at a glance: higher degree in the numerator, the limit is infinite (the sign is decided by the coefficients of the highest-degree terms and by the sign of $x$); equal degrees, it is the ratio of the leading coefficients; higher degree in the denominator, the limit is zero.

It works because for very large $x$ a polynomial behaves like its dominant term. But for $x\to -\infty$ the sign must always be checked: $x^3$ tends to $-\infty$, not to $+\infty$, even though $x^2$ tends to $+\infty$.

$\frac{0}{0}$ with rational functions

If substituting $x_0$ makes numerator and denominator vanish, $(x-x_0)$ is a factor of both. You factorise, cancel and substitute again. The factorisation of Years 1 and 2 (factoring out, difference of squares, difference of cubes, Ruffini) is exactly what is needed here.

The step to keep in mind is the cancelling: you can cancel the factor $(x-x_0)$ because in the limit $x$ approaches $x_0$ without ever reaching it, so the factor is not zero. That is why the limit exists even when the function at $x_0$ is not defined.

$\infty -\infty$ and $\frac{0}{0}$ with roots

Here the technique is rationalisation. You multiply and divide by the conjugate expression, so as to use the difference of squares:

The root disappears and a fraction remains that is handled as in the previous cases. Three typical cases:

• $\sqrt{f(x)}-x$ for $x\to +\infty$: you multiply by $\sqrt{f(x)}+x$;
• $\sqrt{x+a}-b$ in a $\frac{0}{0}$ limit: you multiply by $\sqrt{x+a}+b$;
• difference of two roots: conjugate with the plus sign.

If the form is $\frac{\infty }{\infty }$ with roots, for example $\frac{\sqrt{4x^2+1}}{x+3}$, there is no need to rationalise: you factor out $x^2$ under the root. And here is the commonest trap: $\sqrt{x^2}=|x|$, not $x$. For $x\to +\infty$ it coincides with $x$, for $x\to -\infty$ it equals $-x$. You will see it in exercise 3.

Standard limits: the basic ones

For $\frac{0}{0}$ forms with transcendental functions there are no algebraic manipulations: you use the standard limits, which must be learned by heart. The three basic ones:

The rule for using them: the argument of the function must equal the denominator. When it does not, you multiply and divide by what is missing. For example:

and for a fraction with two functions

because $\cos 2x\to 1$.

The mistakes I see most often

Applying a rule for rational functions to a function with roots. Comparing degrees holds for polynomials. With $\sqrt{x^2+\dots }$ you must first factor out under the root and remember the absolute value.

Forgetting the sign at $-\infty$. For $x\to -\infty$ odd powers are negative and $\sqrt{x^2}=-x$. A limit that equals $2$ at $+\infty$ can equal $-2$ at $-\infty$.

Saying “indeterminate” as the result. The indeterminate form is an obstacle, not an answer. A limit that stops at $\frac{0}{0}$ is an unfinished exercise.

Cancelling terms instead of factors. In $\frac{x^2-4}{x-2}$ you cannot “cancel” the $x$ or the $2$. You factorise the numerator into $(x-2)(x+2)$ and then cancel the whole factor.

Substituting one piece at a time. Reducing one part of the limit to a value first and then continuing with the rest is allowed only when that part is not an indeterminate form. If you substitute before having resolved the obstacle you get a wrong number, and usually it is zero.

Using standard limits with the wrong argument. $\frac{\sin 5x}{x}$ is not $1$: it is $5$.

Three worked exercises

1. $\frac{0}{0}$ with a difference of cubes

Substitution gives $\frac{0}{0}$. We factorise the numerator with the difference of cubes and the denominator with the difference of squares:

Check with a number close to $1$, for example $x=1.01$: the numerator is $1.030301-1=0.030301$, the denominator is $1.0201-1=0.0201$, and the ratio is about $1.507$, close to $\frac{3}{2}$.

2. $\infty -\infty$ with a root

Substitution gives $\infty -\infty$. We multiply and divide by the conjugate:

Now it is $\frac{\infty }{\infty }$. We factor out $x$ in the denominator, knowing that $x>0$ and therefore $\sqrt{x^2}=x$:

Check with $x=100$: $\sqrt{10300}\approx 101.489$, so $101.489-100=1.489$, close to $\frac{3}{2}$.

For $x\to -\infty$ the form is not even indeterminate: $\sqrt{x^2+3x}$ tends to $+\infty$ and $-x$ tends to $+\infty$, so the sum tends to $+\infty$. That is why it pays to recognise the form before starting to rationalise.

3. $\frac{\infty }{\infty }$ with a root, at both infinities

Under the root we factor out $x^2$: $\sqrt{4x^2+1}=|x|\sqrt{4+\frac{1}{x^2}}$. The denominator we write as $x\left(1+\frac{3}{x}\right)$:

For $x\to +\infty$, $|x|=x$ and the limit is $\frac{\sqrt{4}}{1}=2$. For $x\to -\infty$, $|x|=-x$ and the limit is $-\sqrt{4}=-2$.

Check: for $x=1000$ the ratio is $\frac{\sqrt{4000001}}{1003}\approx \frac{2000.0003}{1003}\approx 1.994$. For $x=-1000$ it is $\frac{2000.0003}{-997}\approx -2.006$. The two values are close to $2$ and $-2$: the function has two different horizontal asymptotes, $y=2$ on one side and $y=-2$ on the other, which cannot happen with a rational function.

To practise with more questions there are the guided tests: they are designed for the liceo, but the questions on limits are valid for the technical school too, because the method is the same. If you want the picture of the subject, you will find the rest in maths.

How to prepare for December

The test on limits has a fairly predictable structure: a rational limit, one with a root, one with a standard limit, and almost always a question linked to the asymptote. What I recommend is to do, every evening for a few days, a single limit of each type, and always to write down the form you find after the first substitution. It is a habit of two seconds and it shows straight away in the test: the teacher assesses the reasoning, not just the number.

If the algebra of factorisation is not yet solid, sort it out before limits: half the mistakes I see in December tests start there, not in the limits. In lessons we do exactly this, starting from your own tests and the points where you get stuck. If you want to see where to start, a free introductory call can be booked from contact.