CommeUnJeu · L2 MP

Sequences and series of functions

⌚ ~67 min ▢ 8 blocks ✓ 16 exercises ➣ Prerequisites : Integration on a segment, Limits and continuity in a normed space, Numerical and vector series

Up to now a function and a sequence were two separate objects: a function $f$ takes a real number and returns a value; a sequence $(u_n)$ takes an integer and returns a value. A sequence of functions is both at once: a sequence $(f_n)_{n \in \mathbb{N}}$ whose generic term $f_n$ is itself a function $I \to K$, defined on a common interval $I \subset \mathbb{R}$ with values in $K = \mathbb{R}$ or $\mathbb{C}$. A series of functions is the sequence of partial sums $S_n = \sum_{k=0}^n u_k$ attached to a sequence $(u_n)$ of such functions. Throughout the chapter $I \subset \mathbb{R}$ is an interval (or, when stated, a non-empty subset of $\mathbb{R}$), $K = \mathbb{R}$ or $\mathbb{C}$, and the functions involved are $I \to K$. The notation $\|g\|_\infty = \sup_{x \in I} |g(x)|$ denotes the uniform norm of a bounded function $g$ on $I$, recalled from Normed vector spaces: it measures the maximum gap between two functions. When the underlying set needs to be made explicit we write $\|g\|_{\infty, A} = \sup_{x \in A} |g(x)|$ for $A \subset I$; the subscript is dropped when $A$ is clear from context. The space of continuous functions $I \to K$ is $\mathcal{C}(I, K)$, the space of $k$-times continuously differentiable functions is $\mathcal{C}^k(I, K)$, both recalled from MPSI.
Uniform convergence of $(f_n)$ will repair what pointwise convergence breaks for continuity and integrability on a segment; differentiability needs more --- uniform convergence of the derivatives $(f_n')$ on every segment, plus convergence of $(f_n)$ at a single point --- and the chapter closes with Weierstrass's headline polynomial-approximation theorem. The pedagogical thread is the same throughout: pointwise convergence is too weak; the uniform norm is the right tool, applied to the right sequence. Pointwise convergence transfers nothing on its own --- the limit of continuous functions can be discontinuous, the limit of derivatives can fail to exist, the integral can refuse to follow. Uniform convergence (of the right sequence: of $(f_n)$ for continuity and integration on a segment, of $(f_n')$ for differentiation) repairs every one of those failures.

I Sequences of functions

I.1 Pointwise and uniform convergence

The most natural way to give a meaning to « $f_n \to f$ » freezes the variable $x$: for each $x$, the sequence of numbers $(f_n(x))_n$ must converge to $f(x)$. This is the pointwise (or simple) convergence. It is too lax, as the first Example shows: it allows the limit function to lose properties that every $f_n$ had. The fix is to control the gap uniformly in $x$ --- not « for each $x$ separately » but « one $N$ works for all $x$ at once ». This second notion uses the uniform norm $\|g\|_\infty = \sup_{x \in I} |g(x)|$ recalled in the overview from Normed vector spaces: uniform convergence on $I$ is exactly $\|f_n - f\|_\infty \to 0$. From now on, when we say a function $g$ is bounded on $I$, we mean $\|g\|_\infty < +\infty$ (recalled from MPSI Limits and continuity, extended to $K$-valued functions through the modulus).

Definition — Pointwise convergence

Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions $I \to K$ and $f : I \to K$ a function. The sequence $(f_n)$ converges pointwise (or simply) to $f$ on $I$ when, for every $x \in I$, the sequence of numbers $(f_n(x))_n$ converges to $f(x)$. In quantifiers: $$ \forall x \in I,\ \forall \varepsilon > 0,\ \exists N \in \mathbb{N},\ \forall n \geq N,\ |f_n(x) - f(x)| \leq \varepsilon. $$ The function $f$ is then called the pointwise limit of $(f_n)$ on $I$.

Example — A discontinuous pointwise limit

Take $I = [0, 1]$ and $f_n(x) = x^n$. For $0 \leq x < 1$, $x^n \to 0$; for $x = 1$, $x^n = 1$ for every $n$. So $(f_n)$ converges pointwise on $[0, 1]$ to $$ f(x) = \begin{cases} 0 & \text{if } 0 \leq x < 1,\\ 1 & \text{if } x = 1. \end{cases} $$ Every $f_n$ is continuous on $[0, 1]$; the pointwise limit $f$ is not. Pointwise convergence does not transmit continuity.

Definition — Uniform convergence

Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions $I \to K$ and $f : I \to K$ a function. The sequence $(f_n)$ converges uniformly to $f$ on $I$ when $$ \forall \varepsilon > 0,\ \exists N \in \mathbb{N},\ \forall n \geq N,\ \forall x \in I,\ |f_n(x) - f(x)| \leq \varepsilon. $$ Equivalently: for $n \geq N$ the function $f_n - f$ is bounded on $I$ and $\|f_n - f\|_\infty \leq \varepsilon$, so the condition reads $\|f_n - f\|_\infty \to 0$ once one knows that $f_n - f$ is eventually bounded. The function $f$ is then called the uniform limit of $(f_n)$ on $I$. The order of the quantifiers $\exists N$ and $\forall x$ matters: in uniform convergence the index $N$ depends on $\varepsilon$ but not on $x$; in pointwise convergence $N$ may depend on both $\varepsilon$ and $x$.

Example — Geometric reading of the uniform norm

The Definition admits an immediate geometric translation: $\|f_n - f\|_\infty \leq \varepsilon$ means the graph of $f_n$ lies inside the horizontal strip of half-height $\varepsilon$ around the graph of $f$. Uniform convergence is the demand that, eventually, the whole graph fits in any prescribed strip --- pointwise convergence would only demand that, for each $x$, the value $f_n(x)$ enter the vertical $\varepsilon$-window above $f(x)$, with no link between two different $x$'s.

Proposition — Uniform convergence implies pointwise convergence

If $(f_n)$ converges uniformly to $f$ on $I$, then $(f_n)$ converges \textcolor{colorprop}{pointwise} to $f$ on $I$. The converse is false in general.

This chapter is the theoretical pivot of S1 analysis: it formalises what « $f(x)$ tends to $\ell$ as $x$ tends to $a$ » means and what « $f$ is continuous » means, then proves the seven theorems on which all subsequent analysis (differentiation, integration, asymptotics, integration on a segment, series) leans. Four of these theorems are foundational and load-bearing for the rest of the program --- Heine's sequential characterization, the théorème des valeurs intermédiaires, the théorème des bornes atteintes, and the continuous strictly monotone bijection theorem --- the other three (squeeze, monotone limit, image of an interval) are supporting.
\medskip
Convention. Throughout the chapter, $E \subset \mathbb{R}$ denotes a general subset, $I \subset \mathbb{R}$ an interval with non-empty interior, and $a \in \overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ a real or extended-real point adherent to the function's domain (defined below in the section on Limit at a point: definitions). Limits are stated with large inequalities --- $|x - a| \le \delta$ rather than $|x - a| < \delta$ --- in line with the program's convention « les définitions sont énoncées avec des inégalités larges ». The voisinage-based formulation is mentioned for cultural awareness only; the working definition is the inequality-based one.

We make precise what « $f(x) \to \ell$ as $x \to a$ » means: for every desired closeness to $\ell$, the function eventually stays that close, provided $x$ is close enough to $a$. The four cases (finite or infinite limit at a finite or infinite point) are unified under a single inequality template, with the only difference being how « close to $a$ » and « close to $\ell$ » are spelled out.

Definition — Adherent point

Let $E \subset \mathbb{R}$ and $a \in \overline{\mathbb{R}}$. We say that $a$ is adherent to $E$ if every neighborhood of $a$ meets $E$. For our inequality-based definitions, it is enough to test the symmetric closed intervals $[a - \eta, a + \eta]$ ($\eta > 0$) when $a \in \mathbb{R}$, the tails $[B, +\infty[$ when $a = +\infty$, and the tails $\,]-\infty, B]$ when $a = -\infty$. Hence:

$a \in \mathbb{R}$ is adherent to $E$ if and only if $a \in E$ or $a$ is a limit point of $E$ (every $[a - \eta, a + \eta]$ meets $E$);
$+\infty$ is adherent to $E$ if and only if $E$ is unbounded above;
$-\infty$ is adherent to $E$ if and only if $E$ is unbounded below.

Definition — Finite limit at a finite point

Let $f : E \to \mathbb{R}$, $a \in \mathbb{R}$ adherent to $E$, and $\ell \in \mathbb{R}$. We say that $f$ admits the limit $\ell$ at $a$ if $$ \forall \varepsilon > 0, \ \exists \delta > 0, \ \forall x \in E, \ |x - a| \le \delta \Rightarrow |f(x) - \ell| \le \varepsilon. $$ We then write $f(x) \xrightarrow[x \to a]{} \ell$ or $\lim_{x \to a} f(x) = \ell$.

Skills to practice

Testing pointwise and uniform convergence

I.2 Continuity and the double-limit theorem

Definition — Infinite limit at a finite point

Let $f : E \to \mathbb{R}$ and $a \in \mathbb{R}$ adherent to $E$. We say that $f$ tends to $+\infty$ at $a$ if $$ \forall A \in \mathbb{R}, \ \exists \delta > 0, \ \forall x \in E, \ |x - a| \le \delta \Rightarrow f(x) \ge A. $$ Symmetrically, $f$ tends to $-\infty$ at $a$ if $$ \forall A \in \mathbb{R}, \ \exists \delta > 0, \ \forall x \in E, \ |x - a| \le \delta \Rightarrow f(x) \le A. $$

Definition — Limits at infinity

Let $f : E \to \mathbb{R}$ with $E$ unbounded above (so $+\infty$ is adherent to $E$). Then:

$f$ tends to $\ell \in \mathbb{R}$ at $+\infty$ if $\forall \varepsilon > 0, \exists B \in \mathbb{R}, \forall x \in E, x \ge B \Rightarrow |f(x) - \ell| \le \varepsilon$;
$f$ tends to $+\infty$ at $+\infty$ if $\forall A \in \mathbb{R}, \exists B \in \mathbb{R}, \forall x \in E, x \ge B \Rightarrow f(x) \ge A$;
$f$ tends to $-\infty$ at $+\infty$ if $\forall A \in \mathbb{R}, \exists B \in \mathbb{R}, \forall x \in E, x \ge B \Rightarrow f(x) \le A$.

The symmetric definitions at $-\infty$ (when $E$ is unbounded below) replace the tail $x \ge B$ by $x \le B$.

Example

Show by the definition that $\lim_{x \to 2} (3x + 1) = 7$.

Answer

Fix $\varepsilon > 0$. For $x \in \mathbb{R}$, $$ |(3x + 1) - 7| = |3x - 6| = 3 \, |x - 2|. $$ Choose $\delta = \varepsilon / 3$. Then $|x - 2| \le \delta$ implies $|(3x + 1) - 7| = 3 \, |x - 2| \le 3 \delta = \varepsilon$, which is the required bound.

Example

Show by the definition that $\lim_{x \to 0} 1 / x^2 = +\infty$ (with $E = \mathbb{R}^*$).

Answer

Fix $A \in \mathbb{R}$. We may assume $A > 0$ (otherwise the bound $1/x^2 \ge A$ is trivial since $1/x^2 > 0$). For $x \in \mathbb{R}^*$, $1/x^2 \ge A$ if and only if $x^2 \le 1/A$, i.e. $|x| \le 1 / \sqrt{A}$. Choose $\delta = 1 / \sqrt{A}$. Then $|x - 0| \le \delta$ and $x \ne 0$ imply $1/x^2 \ge A$.

Example

Show by the definition that $\lim_{x \to +\infty} 1/x = 0$ (with $E = \mathbb{R}_+^*$).

Answer

Fix $\varepsilon > 0$. For $x > 0$, $|1/x - 0| = 1/x$, hence $1/x \le \varepsilon$ if and only if $x \ge 1/\varepsilon$. Choose $B = 1 / \varepsilon$. Then $x \ge B \Rightarrow |1/x| \le \varepsilon$.

Example

Compute $\lim_{x \to 0^+} \lfloor x \rfloor$ and $\lim_{x \to 0^-} \lfloor x \rfloor$, and contrast the two.

Answer

For $x \in \,]0, 1[$, $\lfloor x \rfloor = 0$, hence $\lim_{x \to 0^+} \lfloor x \rfloor = 0$. For $x \in \,]-1, 0[$, $\lfloor x \rfloor = -1$, hence $\lim_{x \to 0^-} \lfloor x \rfloor = -1$. The two one-sided limits differ, so $\lfloor \cdot \rfloor$ does not admit a limit at $0$ (cf. P2.4).

Example

Geometric reading of the limit $\lim_{x \to a} f(x) = \ell$.

The graph of $f$ enters the horizontal $\varepsilon$-band $[\ell - \varepsilon, \ell + \varepsilon]$ around $\ell$ as soon as $x$ enters the vertical $\delta$-band $[a - \delta, a + \delta]$ around $a$.

Ex 1

Skills to practice

Applying continuity and the double-limit theorem

I.3 Integration and differentiation

Ex 2 Ex 3

Three structural facts about limits, each one used systematically in the rest of the chapter: uniqueness (the sequential characterization and later sections rely on the limit), the value-at-point coherence imposed by the inclusive convention (a function defined at $a$ admitting a limit there must agree with $f(a)$), and local boundedness (a function with a finite limit cannot blow up nearby). These are the bricks for the algebraic operations on limits in the next section, Operations on limits.

Proposition — Uniqueness of the limit

Let $f : E \to \mathbb{R}$ and $a$ adherent to $E$. If $f$ admits a limit at $a$, that limit is unique.

Proof

Suppose $f$ admits two limits $\ell, \ell' \in \overline{\mathbb{R}}$ at $a$. We prove $\ell = \ell'$ by ruling out the only alternative, $\ell \ne \ell'$.

Both finite. Suppose $\ell, \ell' \in \mathbb{R}$ with $\ell \ne \ell'$. Set $\varepsilon = |\ell - \ell'| / 3 > 0$. The two limit definitions give $\delta_1, \delta_2 > 0$ such that, for $x \in E$ near $a$ (within $\delta = \min(\delta_1, \delta_2)$), $|f(x) - \ell| \le \varepsilon$ and $|f(x) - \ell'| \le \varepsilon$. Such $x$ exists because $a$ is adherent to $E$. By the triangle inequality, $|\ell - \ell'| \le |f(x) - \ell| + |f(x) - \ell'| \le 2 \varepsilon = (2/3) \, |\ell - \ell'|$, hence $|\ell - \ell'| \le 0$ --- contradicting $\ell \ne \ell'$.
One finite, one infinite. Suppose $\ell = +\infty$ and $\ell' \in \mathbb{R}$. Apply the $+\infty$-limit definition with $A = \ell' + 1$: there is a neighborhood of $a$ on which $f \ge \ell' + 1$. Apply the finite-limit definition with $\varepsilon = 1/2$: there is a (possibly smaller) neighborhood on which $|f - \ell'| \le 1/2$, hence $f \le \ell' + 1/2$. The intersection of the two neighborhoods is non-empty (adherence) and forces $\ell' + 1 \le f(x) \le \ell' + 1/2$, contradiction. The case $\ell = -\infty$ is symmetric.
Both infinite, opposite signs. Suppose $\ell = +\infty$ and $\ell' = -\infty$. Apply the $+\infty$-limit definition with $A = 1$ and the $-\infty$-limit definition with $A = -1$: on a common neighborhood of $a$, $f(x) \ge 1$ and $f(x) \le -1$, impossible.

In every case the assumption $\ell \ne \ell'$ leads to a contradiction, hence $\ell = \ell'$.

Inclusive convention vs punctured limits

With our inclusive convention, a function defined at $a$ can have a limit at $a$ only if that limit equals $f(a)$. When one wants to ignore the value at $a$ --- for instance to define a continuous extension (see the later section Continuity at a point) or to describe a removable discontinuity --- one studies the restriction of $f$ to $E \setminus \{a\}$ and speaks of the punctured limit. In what follows, the unqualified word « limit » always means the inclusive limit; the punctured viewpoint will be invoked explicitly when needed.

Proposition — Local boundedness

Let $f : E \to \mathbb{R}$, $a \in \overline{\mathbb{R}}$ adherent to $E$, and assume $f$ admits a finite limit $\ell \in \mathbb{R}$ at $a$. Then $f$ is bounded on some neighborhood of $a$ in $E$:

if $a \in \mathbb{R}$: $\exists \delta > 0, \exists M > 0, \forall x \in E \cap [a - \delta, a + \delta], |f(x)| \le M$;
if $a = +\infty$: $\exists B \in \mathbb{R}, \exists M > 0, \forall x \in E \cap [B, +\infty[, |f(x)| \le M$;
if $a = -\infty$: symmetric tail $\,]-\infty, B]$.

Proof

Apply the limit definition with $\varepsilon = 1$. We get a neighborhood --- $[a - \delta, a + \delta]$ if $a \in \mathbb{R}$, $[B, +\infty[$ if $a = +\infty$, $\,]-\infty, B]$ if $a = -\infty$ --- on which $|f(x) - \ell| \le 1$. By the triangle inequality, $|f(x)| \le |\ell| + 1$ on this neighborhood, hence $f$ is bounded by $M = |\ell| + 1$. Same argument in the three cases.

Skills to practice

Integrating a uniform limit on a segment
Studying the regularity of a limit

II Series of functions

II.1 Modes of convergence

Proposition — Limit and one-sided limits

Let $f : E \to \mathbb{R}$ and $a \in \mathbb{R}$ adherent both to $E \cap \,]a, +\infty[$ and to $E \cap \,]-\infty, a[$. Let $\ell \in \overline{\mathbb{R}}$.

Finite case ($\ell \in \mathbb{R}$). $f$ admits the limit $\ell$ at $a$ if and only if $\lim_{x \to a^+} f(x) = \ell$, $\lim_{x \to a^-} f(x) = \ell$, and (if $a \in E$) $f(a) = \ell$.
Infinite case ($\ell = \pm\infty$). If $a \notin E$, $f$ admits the infinite limit $\ell$ at $a$ if and only if $\lim_{x \to a^+} f(x) = \ell$ and $\lim_{x \to a^-} f(x) = \ell$. If $a \in E$, P2.2 forces the limit to equal $f(a) \in \mathbb{R}$, so under the inclusive convention no infinite limit is possible at a defined point.

The « $f(a) = \ell$ » clause in the finite case is forced by P2.2.

Example

Define $f : \mathbb{R} \to \mathbb{R}$ by $f(x) = x$ for $x \ne 0$ and $f(0) = 1$. Show via P2.2 that $f$ does not admit a limit at $0$.

Answer

Take any sequence $u_n \to 0$ with $u_n \ne 0$. Then $f(u_n) = u_n \to 0$, so by Heine (anticipating T3.1) any limit $\ell$ of $f$ at $0$ would equal $0$. But $0 \in E = \mathbb{R}$, hence by P2.2 any such $\ell$ would also equal $f(0) = 1$. Since $0 \ne 1$, no $\ell$ works, and $f$ has no limit at $0$ in the inclusive sense. The example illustrates that the inclusive convention distinguishes « punctured limit at $0$ » from « inclusive limit at $0$ ».

Example

Show that $f(x) = \sin x$ is bounded on a neighborhood of $0$ without computing the limit, by P2.3 applied with the (obvious) limit $0$.

Answer

We have $\lim_{x \to 0} \sin x = 0$ (lycée fact, recalled in Standard functions). Apply P2.3: there exist $\delta > 0$ and $M > 0$ such that $|\sin x| \le M$ for $|x| \le \delta$. Concretely, with $\varepsilon = 1$ in the proof, $M = 1$ works on any neighborhood, but the point of the example is the structural argument: existence of the bound is automatic from existence of the (finite) limit.

Ex 4 Ex 5 Ex 6

Skills to practice

Proving normal convergence

II.2 Regularity of the sum

Ex 7

Heine's theorem is the single most useful tool of the chapter: limits of functions are completely captured by limits of sequences. The forward direction is direct (apply the limit definition to a convergent sequence); the reverse uses contraposition with a sequence built by negating the limit definition. The theorem unifies the function and sequence chapters and provides the standard recipe for disproving a limit (Method M3.1).

Example

Show that $\sin(1/x)$ has no limit at $0$.

Answer

Take $u_n = 1 / (n \pi)$ and $v_n = 1 / (\pi/2 + 2 n \pi)$ for $n \ge 1$. Both are in $\mathbb{R}^*$ and tend to $0$. Their images: $$ \sin(1 / u_n) = \sin(n \pi) = 0 \to 0, \qquad \sin(1 / v_n) = \sin(\pi/2 + 2 n \pi) = 1 \to 1. $$ The two transferred sequences tend to distinct limits ($0$ and $1$), so by the contrapositive of Heine, $\sin(1/x)$ has no limit at $0$.

Example

(Preview of the composition theorem ; continuity is defined in the section Continuity at a point below.) Show by Heine that if $f \to \ell \in \mathbb{R}$ at $a$ and $g$ is continuous at $\ell$, then $g \circ f \to g(\ell)$ at $a$.

Answer

Let $(u_n)$ be any sequence in the domain of $f$ with $u_n \to a$. By Heine applied to $f$, $f(u_n) \to \ell$. By Heine applied to $g$ (using continuity of $g$ at $\ell$, i.e. $\lim_{y \to \ell} g(y) = g(\ell)$), $g(f(u_n)) \to g(\ell)$. The composed sequence $(g \circ f)(u_n) = g(f(u_n))$ thus tends to $g(\ell)$ for every $u_n \to a$, hence by Heine again $g \circ f \to g(\ell)$ at $a$.

Ex 8 Ex 9

Sums, products, quotients, and compositions of functions with limits behave the way Heine tells us they should: we transfer to sequences and apply the corresponding rules from Suites réelles. The encadrement (squeeze) theorem completes the toolbox, and the théorème de la limite monotone --- a direct corollary of the monotone-sequence theorem --- gives the existence of one-sided limits for monotone functions even without explicit continuity.

Skills to practice

Studying the regularity of a series sum

III Uniform polynomial approximation

III.1 Weierstrass's theorem

Proposition — Algebraic operations on limits

Let $f, g : E \to \mathbb{R}$, $a \in \overline{\mathbb{R}}$ adherent to $E$, and assume $f \to \ell$ and $g \to m$ at $a$ with $\ell, m \in \mathbb{R}$. Then:

$\textcolor{colorprop}{\lambda f + \mu g \to \lambda \ell + \mu m}$ for any $\lambda, \mu \in \mathbb{R}$ (linear combination);
$\textcolor{colorprop}{f g \to \ell m}$ (product);
$\textcolor{colorprop}{f / g \to \ell / m}$ if $m \ne 0$ (quotient; since $g \to m \ne 0$, taking $\varepsilon = |m|/2$ in the limit definition gives $|g(x) - m| \le |m|/2$, hence $|g(x)| \ge |m|/2 > 0$ on a neighborhood of $a$, so $g$ does not vanish there).

The proof transfers to sequences via Heine and applies the corresponding rules for sequences from Suites réelles.

Proposition — Operations with infinite limits

The usual operation rules of P4.1 extend to limits in $\overline{\mathbb{R}}$ whenever the resulting expression has a determined meaning, with the expected sign conditions for products and quotients (e.g.\ $\ell + (+\infty) = +\infty$ for $\ell \in \mathbb{R}$ ; $\ell \cdot (+\infty) = +\infty$ if $\ell > 0$, $-\infty$ if $\ell < 0$ ; $1/(+\infty) = 0$). The four classical indeterminate forms are: $$ \infty - \infty, \quad 0 \cdot \infty, \quad \infty / \infty, \quad 0 / 0. $$ For each indeterminate form, the limit may exist (and take any value, including $\pm\infty$) or may not exist at all; case-by-case analysis is required, typically by factoring the dominant term (M4.1).

Proposition — Passage to the limit of a large inequality

Let $f, g : E \to \mathbb{R}$, $a$ adherent to $E$, and assume $f \le g$ on a neighborhood of $a$. If $\textcolor{colorprop}{f \to \ell}$ and $\textcolor{colorprop}{g \to m}$ at $a$ with $\ell, m \in \overline{\mathbb{R}}$, then $\ell \le m$. The strict inequality is not preserved: $f(x) = -x^2 < 0 = g(x)$ on $\mathbb{R}^*$ but $\lim_{x \to 0} f = \lim_{x \to 0} g = 0$.

Theorem — Squeeze / encadrement

Let $f, g, h : E \to \mathbb{R}$, $a \in \overline{\mathbb{R}}$ adherent to $E$. Assume $f \le g \le h$ on a neighborhood of $a$ in $E$ and $\textcolor{colorprop}{\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = \ell \in \mathbb{R}}$. Then $g$ admits the limit $\ell$ at $a$.

Proof

By Heine, it suffices to show that for every sequence $(u_n)$ in the domain with $u_n \to a$, $g(u_n) \to \ell$. Fix such a $(u_n)$. From a certain rank, $u_n$ lies in the neighborhood where $f \le g \le h$, hence $f(u_n) \le g(u_n) \le h(u_n)$. By Heine applied to $f$ and $h$, $f(u_n) \to \ell$ and $h(u_n) \to \ell$. The squeeze theorem on sequences (Suites réelles) gives $g(u_n) \to \ell$. Heine ($\Leftarrow$) concludes that $g \to \ell$ at $a$.

Proposition — Minoration/majoration with infinite limits

Let $f, g : E \to \mathbb{R}$, $a$ adherent to $E$.

If $f \le g$ on a neighborhood of $a$ and $\textcolor{colorprop}{f \to +\infty}$ at $a$, then $\textcolor{colorprop}{g \to +\infty}$ at $a$.
If $g \le f$ on a neighborhood of $a$ and $\textcolor{colorprop}{f \to -\infty}$ at $a$, then $\textcolor{colorprop}{g \to -\infty}$ at $a$.

This is a one-sided minoration/majoration; the two-sided « squeeze » makes no sense for infinite limits since the upper or lower bound must itself diverge to the same infinity.

Skills to practice

Applying Weierstrass's theorem

III.2 Applications of Weierstrass

Proposition — Composition of limits

Let $f : E \to F \subset \mathbb{R}$, $g : F \to \mathbb{R}$, $a \in \overline{\mathbb{R}}$ adherent to $E$, $b \in \overline{\mathbb{R}}$ adherent to $F$, and $\ell \in \overline{\mathbb{R}}$. If $\textcolor{colorprop}{f \to b}$ at $a$ and $\textcolor{colorprop}{g \to \ell}$ at $b$, then $\textcolor{colorprop}{g \circ f \to \ell}$ at $a$.

Theorem — Théorème de la limite monotone

Let $f : \,]a, b[ \to \mathbb{R}$ be monotone (increasing or decreasing), with $a, b \in \overline{\mathbb{R}}$. Then $f$ admits a (possibly infinite) limit at the left endpoint of the interval (at $a^+$ if $a \in \mathbb{R}$, at $-\infty$ if $a = -\infty$) and at the right endpoint (at $b^-$ if $b \in \mathbb{R}$, at $+\infty$ if $b = +\infty$). Admis (programme; conséquence directe du théorème des suites monotones du cours Suites réelles).

Example

Compute $\lim_{x \to 0^+} x \sin(1/x)$ by squeeze.

Answer

For $x > 0$, $-1 \le \sin(1/x) \le 1$ multiplied by $x > 0$ gives $-x \le x \sin(1/x) \le x$. As $x \to 0^+$, both bounds tend to $0$. By the squeeze theorem, $\lim_{x \to 0^+} x \sin(1/x) = 0$.

Example

Determine $\lim_{x \to 0} (x^2 + x)/x$.

Answer

For $x \ne 0$, $(x^2 + x)/x = x + 1$. By P4.1, $\lim_{x \to 0} (x + 1) = 0 + 1 = 1$. Hence $\lim_{x \to 0} (x^2 + x)/x = 1$.

Example

Compute $\lim_{x \to +\infty} (3 x^2 + 2 x + 1) / (x^2 - 4)$ by factoring.

Answer

Factor $x^2$ from numerator and denominator: $$ \frac{3 x^2 + 2 x + 1}{x^2 - 4} = \frac{x^2 (3 + 2/x + 1/x^2)}{x^2 (1 - 4/x^2)} = \frac{3 + 2/x + 1/x^2}{1 - 4/x^2}. $$ As $x \to +\infty$, $2/x \to 0$, $1/x^2 \to 0$, $4/x^2 \to 0$. By P4.1, the numerator tends to $3$ and the denominator to $1$, hence the quotient to $3 / 1 = 3$.

Skills to practice

Exploiting Weierstrass via continuous functionals