Infinite limits · Topos Uranos

Summary

This lesson studies functions whose values grow or decrease without bound. It completes the common scheme of the definitions of limit with the box that was missing, the one for the value near $\infty$ , and it distinguishes infinite limits, which are a precise way of having no limit, from the other forms of divergence. A single lemma, the reciprocal lemma, reduces infinite limits to limits equal to zero, so that the rules for the sum, the product and the quotient are deduced from those of the algebra of limits; only comparison, the theorem on unbounded monotonic functions and composition require an argument of their own. With them, two results that the earlier lessons left half done are completed: the quotient whose denominator tends to zero and the rational functions whose numerator has the higher degree. The lesson also shows why indeterminate forms are not rules, and it ends with composition with infinite limits.

Learning objectives

Define infinite limits, at a point, on one side and at infinity, with the scheme of the challenge and the response, and distinguish them from the other forms of divergence.
Prove the reciprocal lemma and deduce from it and from comparison the rules for the sum, the product and the quotient with infinite limits.
Recognize indeterminate forms and show with examples that they can give any result.
Compute infinite limits of polynomials, rational functions, trigonometric functions and compositions, and prove the theorem on unbounded monotonic functions.

The missing box

In the lesson on limits as the variable tends to infinity a common scheme was described: every definition of limit is a challenge about the values with a response about the variable, and the only thing that changes is what it means to be near each of them. For the variable, all the forms have already been used: near a number $a$ , on one side of it, or near $\pm\infty$ . For the values, only one: near a number $L$ . This lesson completes the scheme with the remaining form, in which the value is near $\infty$ , that is, $f(x) > N$ , or near $-\infty$ , that is, $f(x) < N$ .

DefinitionInfinite limits

Let $a$ be an accumulation point of the domain $A$ of $f$ . We say that $f$ tends to infinity as $x$ tends to $a$ , and we write $\lim_{x \to a} f(x) = \infty$ , if for every real $N$ there exists a $\delta > 0$ such that $f(x) > N$ for every $x$ in $A$ with $0 < \abs{x - a} < \delta$ . Likewise, $\lim_{x \to a} f(x) = -\infty$ if for every $N$ there exists a $\delta > 0$ such that $f(x) < N$ at those $x$ .

Written with quantifiers, the definition has the usual form; the challenge is now an upper half-plane, that of the points with ordinate greater than $N$ :

\lim_{x \to a} f(x) = \infty \leftrightarrow (\forall N)(\exists \delta > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow f(x) > N\right)

The variants are obtained by changing only the response, as in the earlier lessons: $\lim_{x \to a^+} f(x) = \infty$ if the condition holds for $a < x < a + \delta$ ; $\lim_{x \to \infty} f(x) = \infty$ if there exists an $M$ such that it holds for $x > M$ ; and so on in the other cases. In what follows, $P$ denotes any of the targets of the variable ( $a$ , $a^+$ , $a^-$ , $\infty$ or $-\infty$ ), and “near $P$ ” means in a punctured interval, on one side or on a ray, as the case may be. All the proofs in the lesson use only one common property: two conditions that hold near $P$ hold simultaneously near $P$ , by taking the smaller of two radii or the larger of two bounds. The theorems on finite limits that are cited at an arbitrary target $P$ (uniqueness, local boundedness, preservation of sign, the algebra of limits and the lemmas of the lesson on the algebra of limits) hold at all of them by the lesson on one-sided limits and by the inversion lemma of the preceding lesson, whose correspondence $\delta = \frac{1}{M}$ between the bound and the radius carries infinite limits over in the same way, because it does not depend on the form of the challenge.

It is worth insisting on what the notation says. If $\lim_{x \to a} f(x) = \infty$ , the function has no limit at $a$ in the sense of the lesson on the limit of functions of a real variable, because it is not bounded near $a$ , and local boundedness says that a function with a limit is. The equality with $\infty$ describes a precise way of having no limit. We say that $f$ diverges at $P$ if it has no finite limit there; tending to $\infty$ or to $-\infty$ are two ways of diverging, but not the only ones. The function $\sin \frac{1}{x}$ diverges at $0$ and is bounded; the function $\frac{1}{x}\sin\frac{1}{x}$ diverges at $0$ , is not bounded near $0$ and, nevertheless, tends neither to $\infty$ nor to $-\infty$ , because it vanishes at points as close to $0$ as we like.

PropositionThe reciprocals of

x^2

and of

x

at zero

$\lim_{x \to 0} \frac{1}{x^2} = \infty$ . On the other hand, $\lim_{x \to 0^+} \frac{1}{x} = \infty$ and $\lim_{x \to 0^-} \frac{1}{x} = -\infty$ , so that $\frac{1}{x}$ tends neither to $\infty$ nor to $-\infty$ as $x$ tends to $0$ . The response to the challenge $N$ is the square root of the reciprocal of $N$ . In the figure, the upper half-plane is the challenge; the strip adjacent to $0$ , the response.

Proof

$\delta = \dato{d}{\frac{1}{\sqrt{N}}}$
Let $N$ be a real number; it suffices to consider $N > 0$ , because a response to a larger challenge responds to all the smaller ones. We take as the response the square root of its reciprocal.
$0 < x^2 < \frac{1}{N} \rightarrow \resaltar{\frac{1}{x^2} > N}$
If $0 < \abs{x} < \delta$ , the square of $x$ is positive and less than $\frac{1}{N}$ ; taking reciprocals of positive numbers reverses the inequality.
$0 < x < \frac{1}{N} \rightarrow \frac{1}{x} > N, \qquad -\frac{1}{N} < x < 0 \rightarrow \frac{1}{x} < -N$
For $\frac{1}{x}$ from the right, the proposition on the reciprocal from the preceding lesson gives the response $\delta = \frac{1}{N}$ . From the left, $\frac{1}{x}$ is the opposite of $\frac{1}{-x}$ , and $-x$ lies to the right of $0$ : the values are less than $-N$ .
$\lim_{x \to 0^+} \frac{1}{x} = \infty \neq -\infty = \lim_{x \to 0^-} \frac{1}{x}$
In every punctured interval with center $0$ there are values of $\frac{1}{x}$ greater than $1$ and less than $-1$ : no $\delta$ responds to the challenge $N = 1$ for $\infty$ , nor to the challenge $N = -1$ for $-\infty$ .