Limes fvnctionvm vnivs variabilis realis

Svmmarivm

Haec lectio in definitionem accvratam convertit notionem valores fvnctionis ad nvmervm accedere, cvm variabilis ad pvnctvm accedit. Ostendit cvr notio intvitiva non svfficiat, pvncta accvmvlationis introdvcit, qvae indicant vbi de limite qvaerere sensvm habeat, et definitionem per epsilon et delta envntiat, qvae vt provocatio eivsqve responsio legitvr. Definitionem eivsqve negationem vt envntiata qvantificata scribit, et demonstrat limitem, si exsistit, vnicvm esse, ex solis valoribvs pvncto propinqvis pendere, et fvnctionem cogere vt prope id pvnctvm circvmscripta maneat signvmqve servet. Problemata delta constrvvnt pro fvnctionibvs polynomialibvs, radicibvs et qvotientibvs, ostendvnt qvosdam limites non exsistere, et maximvm delta qvaervnt qvod dato epsilon respondet.

Proposita discendi

Notionem limitis fvnctionis in pvncto explicare eamqve a valore fvnctionis in eodem pvncto distingvere.
Definitionem limitis per epsilon et delta envntiare, postvlatvm ivstificare vt pvnctvm sit pvnctvm accvmvlationis dominii, et definitionem eivsqve negationem qvantificatoribvs scribere.
Ex definitione valorem limitis fvnctionvm algebraicarvm simplicivm demonstrare, pro vnoqvoqve epsilon delta constrvendo.
Vnicitatem limitis eivsqve proprietates locales demonstrare, et negationem definitionis adhibere ad ostendendvm limitem aliqvem non exsistere.

Notio limitis

Algebra et geometria stvdent qvem valorem fvnctio in vnoqvoqve pvncto habeat; calcvlvs praeterea stvdet qvid valoribvs eivs fiat, cvm variabilis ad pvnctvm accedit, etiamsi in eo pvncto fvnctio definita non est. Necessitas antiqva est. Declivitas rectae qvae dvo pvncta cvrvae ivngit qvotiente differentiarvm compvtatvr; declivitas rectae tangentis esse deberet valor eivs qvotientis, cvm dvo pvncta confvndvntvr, sed tvm qvotiens est $\frac{0}{0}$ , qvi nvmervs non est. Eodem modo velocitas media corporis moti est qvotiens spatii confecti et temporis consvmpti, et velocitas in instanti est id qvo is qvotiens tandem pervenit, cvm tempvs consvmptvm ad nihilvm redigitvr. Vtroqve in casv de valoribvs fvnctionis qvaeritvr prope pvnctvm vbi definita non est.

Exemplvm simplex rem ostendit. Fvnctio qvae seqvitvr in $x = 1$ definita non est, qvia ibi denominator eivs evanescit; tamen valores eivs in pvnctis prope $1$ sitis prope $2$ svnt:

f(x) = \frac{x^2 - 1}{x - 1}, \qquad f(0{,}9) = 1{,}9, \quad f(0{,}99) = 1{,}99, \quad f(1{,}01) = 2{,}01, \quad f(1{,}1) = 2{,}1

Ratio est qvod, pro $x \neq 1$ , nvmerator in factores $(x - 1)(x + 1)$ resolvitvr et qvotiens valet $x + 1$ ; delineatio ipsivs $f$ est recta $y = x + 1$ cvm foramine in pvncto $(1, 2)$ . Dicitvr igitvr limitem ipsivs $f$ , cvm $x$ ad $1$ tendit, esse $2$ , et scribitvr:

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2

Atqvi sententia «valores ad $2$ accedvnt» vt definitio non svfficit, dvabvs de cavsis. Prima est qvod accedere non svfficit: cvm $x$ a $0{,}9$ ad $0{,}99$ transit, valor ipsivs $f$ etiam ad $3$ accedit, qvia $1{,}99$ propivs ad $3$ est qvam $1{,}9$ , et tamen $3$ limes non est. Qvod nvmervm $2$ distingvit est qvod valores ipsivs $f$ ei qvam proxime libet accedvnt, non solvm propivs qvam antea. Altera cavsa est qvod tabvla nvmervm finitvm pvnctorvm examinat, et fvnctio inter ea aliter se gerere potest: fvnctio $\sin\left(\frac{\pi}{x}\right)$ valet $0$ in $x = 0{,}1$ , in $x = 0{,}01$ et in omnibvs pvnctis formae $\frac{1}{n}$ , et tamen limitem in $0$ non habet, qvia inter ea pvncta inter $-1$ et $1$ oscillat; problema propositvm 3 id pro $\sin\left(\frac{1}{x}\right)$ eodem argvmento demonstrat. Qvod si ita est, definitio de omnibvs pvnctis propinqvis simvl loqvi debet et nvmeris figere qvid significet «qvam proxime libet». Instrvmenta ad hoc svnt valor absolvtvs, qvi distantias metitvr, et inaeqvalitas triangvlaris, in lectione de recta reali tractata.

Pvncta accvmvlationis

Anteqvam limes definiatvr, accvrativs dicendvm est in qvibvs pvnctis de eo qvaerere sensvm habeat. Pvnctvm $a$ ad dominivm fvnctionis pertinere non necesse est, vt exemplvm svpra ostendit; sed pvncta dominii esse debent qvam proxime libet ad $a$ , qvia aliter nvlli valores fvnctionis observandi essent.

DefinitioPvnctvm accvmvlationis

Sit $A$ copia nvmerorvm realivm. Nvmervs realis $a$ est pvnctvm accvmvlationis copiae $A$ , si omne intervallvm apertvm centrvm $a$ habens aliqvod elementvm copiae $A$ a $a$ diversvm continet:

(\forall \delta > 0)(\exists x \in A)(0 < \abs{x - a} < \delta)

Elementvm copiae $A$ qvod pvnctvm accvmvlationis copiae $A$ non est pvnctvm solitarivm copiae $A$ appellatvr.

Exempli gratia, $1$ est pvnctvm accvmvlationis copiae $\mathbb{R} \setminus \{1\}$ , qvae est dominivm fvnctionis exempli, qvamvis ad eam non pertineat. Omne pvnctvm intervalli clavsi $\icc{0}{1}$ est pvnctvm accvmvlationis intervalli aperti $\ioo{0}{1}$ , terminis inclvsis. Nvmervs $0$ est pvnctvm accvmvlationis copiae nvmerorvm $\frac{1}{n}$ cvm $n$ natvrali, qvia per proprietatem Archimedeam pro omni $\delta > 0$ est nvmervs natvralis $n$ cvm $\frac{1}{n} < \delta$ . Contra, copia $\mathbb{N}$ nvlla pvncta accvmvlationis habet: intervallvm centri $a$ et radii $\frac{1}{2}$ ad svmmvm vnvm nvmervm natvralem $m$ continet; si $m = a$ , nvllvm nvmervm natvralem a $a$ diversvm continet, et si $m \neq a$ , intervallvm radii $\abs{m - a}$ iam nvllvm continet. Condicio $0 < \abs{x - a}$ essentialis est; sine ea omne elementvm copiae $A$ esset pvnctvm accvmvlationis copiae $A$ , qvia qvodqve intervallvm centrvm $a$ habens ipsvm $a$ continet.

Definitio limitis

His notionibvs definitio dari potest. Sententia « $f(x)$ est qvam proxime libet ad $L$ » sic convertitvr: pro vnaqvaqve distantia $\varepsilon > 0$ , qvamvis parva, valores ipsivs $f$ minvs qvam $\varepsilon$ ab $L$ distant, dvmmodo $x$ satis prope $a$ sit, id est minvs qvam certa distantia $\delta > 0$ , neqve aeqvalis $a$ .

DefinitioLimes fvnctionis in pvncto

Sint $f$ fvnctio realis in copia $A$ nvmerorvm realivm definita, $a$ pvnctvm accvmvlationis copiae $A$ et $L$ nvmervs realis. Dicitvr $L$ esse limes ipsivs $f$ , cvm $x$ ad $a$ tendit, et scribitvr $\lim_{x \to a} f(x) = L$ , si pro omni $\varepsilon > 0$ exsistit $\delta > 0$ tale vt omne $x$ copiae $A$ qvod condicioni $0 < \abs{x - a} < \delta$ satisfacit etiam condicioni $\abs{f(x) - L} < \varepsilon$ satisfaciat:

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

Geometrice condicio $\abs{f(x) - L} < \varepsilon$ dicit pvnctvm $(x, f(x))$ delineationis in fascia horizontali esse inter rectas $y = L - \varepsilon$ et $y = L + \varepsilon$ comprehensa; condicio $0 < \abs{x - a} < \delta$ dicit $x$ in intervallo centri $a$ et radii $\delta$ esse, neqve esse eivs centrvm. Definitio vt lvdvs inter dvas personas legi potest: prima proponit $\varepsilon$ , radivm fasciae circa $L$ , tam angvstae qvam velit; altera respondere debet $\delta$ tali vt portio delineationis svpra intervallvm perforatvm radii $\delta$ sita intra fasciam maneat. Limes est $L$ , si altera persona omnibvs provocationibvs respondere potest.

ExemplvmProvocatio et responsio

Fvnctio $g$ valet $x^2$ in omni $x \neq 1$ , et in $x = 1$ ei de indvstria tribvitvr valor qvi cvm ceteris non congrvit: $g(1) = \frac{5}{2}$ . Videtvr qvomodo dvabvs provocationibvs respondeatvr pro limite $L = 1$ in pvncto $a = 1$ .

Demonstratio

$(\forall x \neq 1)\left(g(x) = x^2\right), \qquad g(1) = \frac{5}{2}$
Fvnctio in tota recta definita est. Prope $1$ valores eivs propinqvi svnt nvmero $1$ , praeter valorem in $x = 1$ , qvi seorsvm manet.
$\abs{g(x) - 1} < \frac{4}{5} \leftrightarrow \frac{1}{5} < g(x) < \frac{9}{5}$
Prima provocatio: $\varepsilon = \frac{4}{5}$ . Valores ipsivs $g$ manere debent in fascia qvae a $\frac{1}{5}$ ad $\frac{9}{5}$ patet.
$\delta = \min\left(1 - \sqrt{\frac{1}{5}}, \sqrt{\frac{9}{5}} - 1\right) = \resaltar{\sqrt{\frac{9}{5}} - 1}$
Pro $x$ positivis a $1$ diversis condicio aeqvivalet ei qvod $x$ inter $\sqrt{\frac{1}{5}}$ et $\sqrt{\frac{9}{5}}$ est. Intervallvm respectv $1$ symmetricvm non est: terminvs dexter propior est. Respondetvr minore dvarvm distantiarvm.
$0 < \abs{x - 1} < \delta \rightarrow \abs{g(x) - 1} < \frac{4}{5}$
Omne $x$ qvod a $1$ minvs qvam $\delta$ distat, neqve est $1$ , condicioni satisfacit: portio delineationis svpra intervallvm perforatvm in fascia manet. Pvnctvm $(1, g(1))$ extra fasciam est, sed nihil refert, qvia definitio $x = 1$ exclvdit.
$\varepsilon = \frac{2}{5} \rightarrow \delta = \resaltar{\sqrt{\frac{7}{5}} - 1}$
Provocatio severior, $\varepsilon = \frac{2}{5}$ , minorem responsionem exigit. Generatim pro $0 < \varepsilon < 1$ svfficit $\delta = \sqrt{1 + \varepsilon} - 1$ , qvod tam parvvm fit qvam libet, cvm $\varepsilon$ tale est: $\delta$ ab $\varepsilon$ pendet.

Exemplvm tres notas definitionis ostendit qvas prae ocvlis habere convenit. Primvm, $\delta$ ab $\varepsilon$ pendet, et generatim etiam a pvncto $a$ ; si $\delta$ provocationi respondet, qvilibet nvmervs positivvs eo minor etiam respondet, ita vt vnvm invenire svfficiat, neqve maximvm reqviratvr. Deinde, valor fvnctionis in $a$ , si exsistit, nihil confert: condicio $0 < \abs{x - a}$ evm exclvdit, ideoqve limes ipsivs $g$ in $1$ est $1$ , qvamvis $g(1)$ valeat $\frac{5}{2}$ . Postremo, sola pvncta $x$ dominii nvmerantvr; cetera non examinantvr.

ExemplvmFvnctio linearis

Demonstratvr $\lim_{x \to 2} (3x - 1) = 5$ . Dato $\varepsilon > 0$ , efficiendvm est vt $\abs{(3x - 1) - 5} < \varepsilon$ . Expressio simplificatvr et relationem svam cvm $\abs{x - 2}$ ostendit:

\abs{(3x - 1) - 5} = \abs{3x - 6} = 3\abs{x - 2}

Svfficit igitvr vt $3\abs{x - 2} < \varepsilon$ , id est vt $\abs{x - 2} < \frac{\varepsilon}{3}$ . Respondetvr $\delta = \frac{\varepsilon}{3}$ : si $0 < \abs{x - 2} < \frac{\varepsilon}{3}$ , tvm $\abs{(3x - 1) - 5} < 3 \cdot \frac{\varepsilon}{3} = \varepsilon$ . Eodem modo pro fvnctione constanti $c$ qvodlibet $\delta$ respondet, qvia $\abs{c - c} = 0 < \varepsilon$ ; et pro fvnctione identica respondet $\delta = \varepsilon$ , qvia condicio postvlata, $\abs{x - a} < \varepsilon$ , eadem est ac qvae svpponitvr. Ergo:

\lim_{x \to a} c = c, \qquad \lim_{x \to a} x = a

Methodvs exempli est generalis: $\abs{f(x) - L}$ ita scribitvr vt factor $\abs{x - a}$ appareat, et reliqvvm circvmscribitvr. Cvm reliqvvm constans non est, circvmscribitvr restringendo antea $\delta$ ad valorem fixvm, exempli gratia $\delta \leq 1$ ; haec est ars exempli qvod seqvitvr.

ExemplvmQvadratvm

Demonstratvr $\lim_{x \to 3} x^2 = 9$ . Sit $\varepsilon > 0$ .

Demonstratio

$\abs{x^2 - 9} = \abs{x - 3} \cdot \dato{f}{\abs{x + 3}}$
Differentia qvadratorvm in factores resolvitvr; apparet factor $\abs{x - 3}$ , qvi parvvs reddetvr, mvltiplicatvs per factorem qvi ab $x$ pendet.
$\abs{x - 3} < 1 \rightarrow 5 < x + 3 < 7 \rightarrow \dato{c}{\abs{x + 3} < 7}$
Si postvlatvr $\abs{x - 3} < 1$ , tvm $x$ inter $2$ et $4$ est, et factor secvndvs per $7$ circvmscribitvr.
$\delta = \min\left(1, \frac{\varepsilon}{7}\right)$
Hac restrictione posita, svfficit praeterea vt $7\abs{x - 3} < \varepsilon$ . Eligitvr $\delta$ vt minor dvorvm nvmerorvm, vt ambae condiciones impleantvr.
$\abs{x^2 - 9} = \abs{x - 3}\abs{x + 3} < \frac{\varepsilon}{7} \cdot 7 = \resaltar{\varepsilon}$
Si $0 < \abs{x - 3} < \delta$ , tvm $\abs{x - 3} < 1$ et $\abs{x - 3} < \frac{\varepsilon}{7}$ ; inde seqvitvr inaeqvalitas postvlata.

Limes initii lectionis eodem modo demonstratvr, cvm observatione qva saepe vtemvr: cvm definitio pvnctvm $a$ non examinet, expressionem ipsivs $f$ in pvnctis a $a$ diversis simplificare licet. Pro $x \neq 1$ qvotiens $\frac{x^2 - 1}{x - 1}$ valet $x + 1$ , et tvm $\abs{f(x) - 2} = \abs{x - 1}$ ; respondet $\delta = \varepsilon$ . Factorem $x - 1$ tollere legitimvm est propterea qvod $x$ in definitione nvmqvam valet $1$ .

Definitio vt envntiatvm qvantificatvm

Definitio limitis est envntiatvm tribvs qvantificatoribvs alternis constans: pro omni $\varepsilon$ exsistit $\delta$ , pro omni $x$ . Strvctvra eivs logica est envntiatorvm qvae logica praedicatorvm stvdet, et eam tractare scire dimidivm laboris cvm limitibvs est. Dvo consectaria ex ea flvvnt.

Prima est qvod ordo qvantificatorvm refert. In definitione $\delta$ eligitvr postqvam $\varepsilon$ cognitvm est, ideoqve ab eo pendere potest. Si primvm $(\exists \delta > 0)$ et deinde $(\forall \varepsilon > 0)$ scriberetvr, vnvm idemqve $\delta$ omnibvs $\varepsilon$ simvl servire deberet, qvae assertio mvlto fortior est, qvam problema propositvm 4 notis propriis definire ivbet.

Altera est qvod negatio definitionis regvlis negationis qvantificatorvm obtinetvr: negatio «pro omni» est «exsistit aliqvod... qvod non», et negatio «exsistit» est «pro omni... non». Sic obtinetvr qvid significet $L$ limitem non esse.

PropositioNegatio definitionis limitis

Sint $f$ fvnctio in $A$ definita, $a$ pvnctvm accvmvlationis copiae $A$ et $L$ nvmervs realis. Nvmervs $L$ non est limes ipsivs $f$ , cvm $x$ ad $a$ tendit, si et solvm si exsistit $\varepsilon > 0$ tale vt pro omni $\delta > 0$ sit aliqvod $x$ copiae $A$ cvm $0 < \abs{x - a} < \delta$ et $\abs{f(x) - L} \geq \varepsilon$ . Negatio introrsvm fertvr, qvantificator post qvantificatorem, donec implicationem attingat.

Demonstratio

$\neg (\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)$
Tota definitio negatvr.
$(\exists \varepsilon > 0) \resaltar{\neg} (\exists \delta > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)$
Negatio envntiati vniversalis est exsistentia casvs in qvo deficit.
$(\exists \varepsilon > 0)(\forall \delta > 0) \resaltar{\neg} (\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)$
Negatio envntiati exsistentialis est omnes casvs deficere.
$(\exists \varepsilon > 0)(\forall \delta > 0)(\exists x \in A) \resaltar{\neg} \left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)$
Rvrsvs envntiatvm vniversale negatvr.
$(\exists \varepsilon > 0)(\forall \delta > 0)(\exists x \in A)\left(0 < \abs{x - a} < \delta \wedge \resaltar{\abs{f(x) - L} \geq \varepsilon}\right)$
Implicatio falsa est si et solvm si antecedens eivs vervm est et conseqvens falsvm; et negatio ipsivs $\abs{f(x) - L} < \varepsilon$ est $\abs{f(x) - L} \geq \varepsilon$ , per trichotomiam.

Lvdi verbis, $L$ limes non est, si prima persona provocationem habet cvi nvlla responsio svfficit: qvamvis parvvm sit $\delta$ propositvm, est pvnctvm intervalli perforati cvivs valor e fascia exit. Ad ostendendvm fvnctionem in $a$ limitem non habere, ostendendvm est nvllvm nvmervm realem $L$ limitem esse, vt fit in problemate solvto 4.

Proprietates fvndamentales

Notatio $\lim_{x \to a} f(x) = L$ de limite vno loqvitvr, qvasi vnicvs nvmervs sit. Id demonstrationem exigit: dvos nvmeros diversos definitioni simvl satisfacere non posse. Hic intervenit postvlatvm vt $a$ sit pvnctvm accvmvlationis dominii. Qvod si non esset, exsisteret $\delta_0 > 0$ tale vt nvllvm $x$ dominii condicioni $0 < \abs{x - a} < \delta_0$ satisfaceret; semper eo $\delta_0$ respondendo, implicatio definitionis pro omni $L$ vera esset, qvia antecedens eivs nvmqvam impleretvr, et omnis nvmervs limes esset.

TheoremaVnicitas limitis

Sint $f$ fvnctio in $A$ definita et $a$ pvnctvm accvmvlationis copiae $A$ . Si $\lim_{x \to a} f(x) = L$ et $\lim_{x \to a} f(x) = L'$ , tvm $L = L'$ . Ratiocinatio fit per redvctionem ad absvrdvm: ponitvr limites diversos esse, exempli gratia $L < L'$ , et eligitvr $\varepsilon$ tam parvvm vt fasciae radii $\varepsilon$ circa $L$ et circa $L'$ se non tangant.

Demonstratio

$\dato{e}{\varepsilon = \frac{L' - L}{2} > 0}$
Ponamvs $L < L'$ , et sit $\varepsilon$ dimidivm distantiae inter vtrvmqve; positivvm est.
$0 < \abs{x - a} < \delta_1 \rightarrow \abs{f(x) - L} < \varepsilon, \qquad 0 < \abs{x - a} < \delta_2 \rightarrow \abs{f(x) - L'} < \varepsilon$
Fasciae radii $\varepsilon$ circa $L$ et circa $L'$ nvlla pvncta commvnia habent, et definitio ad vtrvmqve limitem eo $\varepsilon$ adhibetvr: obtinentvr dvae responsiones, $\delta_1$ et $\delta_2$ , qvae generatim diversae svnt.
$\delta = \min(\delta_1, \delta_2), \quad (\exists x \in A)\left(0 < \abs{x - a} < \delta\right)$
Vt ambae condiciones simvl valeant, minor dvorvm radiorvm svmitvr. Cvm $a$ sit pvnctvm accvmvlationis copiae $A$ , intervallvm perforatvm radii $\delta$ aliqvod $x$ dominii continet.
$\dato{d}{\abs{f(x) - L} < \varepsilon \wedge \abs{f(x) - L'} < \varepsilon}$
Pro eo $x$ ambae inaeqvalitates valent: valor $f(x)$ in ambabvs fasciis simvl esset.
$L' - L \leq \abs{L' - f(x)} + \abs{f(x) - L} < 2\varepsilon = \resaltar{L' - L}$
Per inaeqvalitatem triangvlarem distantia inter $L$ et $L'$ minor esset qvam $2\varepsilon$ , qvae est ipsa ea distantia: contradictio. Ergo esse non potest $L < L'$ ; eodem modo esse non potest $L' < L$ , et per trichotomiam $L = L'$ .

Lemma qvod seqvitvr observationem formaliter exprimit qva qvotientem $\frac{x^2 - 1}{x - 1}$ simplificare licvit: limes in $a$ ex solis valoribvs fvnctionis prope $a$ pendet, non ex eo qvem in $a$ svmit neqve ex iis procvl ab eo.

LemmaLimes ex solis valoribvs propinqvis pendet

Sint $f$ et $g$ fvnctiones in $A$ definitae, $a$ pvnctvm accvmvlationis copiae $A$ et $r > 0$ tale vt $f(x) = g(x)$ pro omni $x$ copiae $A$ cvm $0 < \abs{x - a} < r$ . Si $\lim_{x \to a} f(x) = L$ , tvm $\lim_{x \to a} g(x) = L$ . Svfficit responsionem svmere qvae pro $f$ valet et, si opvs est, eam vsqve ad $r$ minvere.

Demonstratio

$\dato{f}{0 < \abs{x - a} < \delta_1 \rightarrow \abs{f(x) - L} < \varepsilon}$
Sit $\varepsilon > 0$ . Ex hypothesi exsistit responsio $\delta_1$ pro $f$ .
$\delta = \min(\delta_1, r)$
Minor svmitvr inter eam responsionem et radivm in qvo fvnctiones congrvvnt.
$\abs{g(x) - L} = \abs{\resaltar{f(x)} - L} < \varepsilon$
Si $x$ ad $A$ pertinet et $0 < \abs{x - a} < \delta$ , tvm $g(x) = f(x)$ , qvia $\delta \leq r$ , et $f(x)$ in fascia est, qvia $\delta \leq \delta_1$ .

Dvae proprietates qvae seqvvntvr dicvnt limitem fvnctioni prope pvnctvm condiciones imponere. Prima eam cogit circvmscriptam manere; altera, signvm limitis habere, cvm is a nihilo differt. Ambae obtinentvr definitionem adhibendo cvm $\varepsilon$ antea electo, et instrvmenta ervnt ad regvlas prodvcti et qvotientis demonstrandas in lectione de algebra limitvm.

PropositioCircvmscriptio localis et conservatio signi

Sint $f$ fvnctio in $A$ definita, $a$ pvnctvm accvmvlationis copiae $A$ et $\lim_{x \to a} f(x) = L$ . Tvm exsistit $\delta > 0$ tale vt pro omni $x$ copiae $A$ cvm $0 < \abs{x - a} < \delta$ valeat $\abs{f(x)} < \abs{L} + 1$ . Si praeterea $L \neq 0$ , exsistit $\delta > 0$ tale vt pro iisdem $x$ ipsvm $f(x)$ signvm ipsivs $L$ habeat et $\abs{f(x)} > \frac{\abs{L}}{2}$ . Vtraqve assertio est definitio ad provocationem certam adhibita: $\varepsilon = 1$ pro prima et $\varepsilon = \frac{\abs{L}}{2}$ pro altera.

Demonstratio

$0 < \abs{x - a} < \delta \rightarrow \dato{u}{\abs{f(x) - L} < 1}$
Cvm $\varepsilon = 1$ definitio dat $\delta$ tale vt valores ipsivs $f$ ab $L$ minvs qvam $1$ distent.
$\abs{f(x)} \leq \abs{f(x) - L} + \abs{L} < \resaltar{\abs{L} + 1}$
Per inaeqvalitatem triangvlarem valor absolvtvs ipsivs $f(x)$ non excedit valorem ipsivs $L$ avctvm ea distantia.
$\dato{m}{L - \frac{\abs{L}}{2} < f(x) < L + \frac{\abs{L}}{2}}$
Si $L \neq 0$ , cvm $\varepsilon = \frac{\abs{L}}{2}$ , qvod positivvm est, definitio alivd $\delta$ dat, et $f(x)$ in intervallo centri $L$ et radii $\frac{\abs{L}}{2}$ manet.
$L > 0 \rightarrow f(x) > \resaltar{\frac{L}{2}} > 0, \qquad L < 0 \rightarrow f(x) < \resaltar{\frac{L}{2}} < 0$
Si $L > 0$ , terminvs sinister est $\frac{L}{2}$ , qvi positivvs est; si $L < 0$ , terminvs dexter est $\frac{L}{2}$ , qvi negativvs est. Vtroqve in casv $f(x)$ signvm ipsivs $L$ habet et longivs a $0$ abest qvam $\frac{\abs{L}}{2}$ .

Definitione valor cvivslibet limitis demonstrari potest, sed labor cvm complexitate fvnctionis crescit. Lectio de algebra limitvm ex hac regvlas demonstrat pro svmma, prodvcto, qvotiente, potentiis et radicibvs, qvae plerosqve limites compvtare sinvnt, qvin semper rvrsvs $\delta$ qvaerendvm sit.

Problemata solvta

Problemata difficvltate crescvnt: tria prima $\delta$ constrvvnt pro polynomio, radice et qvotiente, qvodqve svo modo factorem circvmscribendi qvi $\abs{x - a}$ comitatvr; qvartvm negatione definitionis ostendit limitem aliqvem non exsistere; vltimvm maximvm $\delta$ qvaerit qvod datae provocationi respondet.

Problema solvtvm 1Polynomivm secvndi gradvs

Ex definitione demonstrare $\lim_{x \to 2} (x^2 + x) = 6$ .

Solvtio

$\abs{(x^2 + x) - 6} = \abs{x - 2} \cdot \dato{f}{\abs{x + 3}}$
Sit $\varepsilon > 0$ . Cvm $2$ sit radix polynomii $x^2 + x - 6$ , id factorem $x - 2$ habet.
$\abs{x - 2} < 1 \rightarrow 4 < x + 3 < 6 \rightarrow \dato{c}{\abs{x + 3} < 6}$
Radivs antea ad $1$ restringitvr: tvm $x$ inter $1$ et $3$ est, et factor secvndvs circvmscribitvr.
$\delta = \min\left(1, \frac{\varepsilon}{6}\right)$
Hac restrictione posita svfficit vt $6\abs{x - 2} < \varepsilon$ ; eligitvr minor dvorvm radiorvm.
$\abs{(x^2 + x) - 6} = \abs{x - 2}\abs{x + 3} < \frac{\varepsilon}{6} \cdot 6 = \resaltar{\varepsilon}$
Si $0 < \abs{x - 2} < \delta$ , ambae condiciones implentvr, et inaeqvalitas postvlata seqvitvr.

Problema solvtvm 2Radix qvadrata

Ex definitione demonstrare $\lim_{x \to 4} \sqrt{x} = 2$ , vbi fvnctio in $\ico{0}{\infty}$ definita est.

Solvtio

$\sqrt{x} - 2 = \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{\sqrt{x} + 2} = \dato{q}{\frac{x - 4}{\sqrt{x} + 2}}$
Sit $\varepsilon > 0$ . Vt $x - 4$ appareat, mvltiplicatvr et dividitvr per expressionem conivgatam, $\sqrt{x} + 2$ , qvae positiva est.
$\abs{\sqrt{x} - 2} = \frac{\abs{x - 4}}{\sqrt{x} + 2} \leq \dato{c}{\frac{\abs{x - 4}}{2}}$
Denominator saltem $2$ est, qvia $\sqrt{x} \geq 0$ ; hac vice radivm restringere non opvs est, qvia margo in toto dominio valet.
$\delta = 2\varepsilon$
Svfficit igitvr vt $\frac{\abs{x - 4}}{2} < \varepsilon$ .
$\abs{\sqrt{x} - 2} \leq \frac{\abs{x - 4}}{2} < \frac{2\varepsilon}{2} = \resaltar{\varepsilon}$
Si $x \geq 0$ et $0 < \abs{x - 4} < 2\varepsilon$ , inaeqvalitas postvlata obtinetvr. Sola pvncta $x$ dominii examinantvr: negativa, in qvibvs radix definita non est, non nvmerantvr, qvamvis prope $4$ sint, si $\varepsilon$ magnvm est.

Problema solvtvm 3Qvotiens

Ex definitione demonstrare $\lim_{x \to 3} \frac{1}{x} = \frac{1}{3}$ , vbi fvnctio in $\mathbb{R} \setminus \{0\}$ definita est.

Solvtio

$\abs{\frac{1}{x} - \frac{1}{3}} = \abs{\frac{3 - x}{3x}} = \frac{\abs{x - 3}}{\dato{d}{3\abs{x}}}$
Sit $\varepsilon > 0$ . Differentia ad commvnem denominatorem redvcitvr.
$\abs{x - 3} < 1 \rightarrow x > 2 \rightarrow \dato{c}{3\abs{x} > 6}$
Nvnc factor qvi $\abs{x - 3}$ comitatvr in denominatore est, et prohibendvm est ne is ad $0$ accedat. Si $\abs{x - 3} < 1$ , tvm $x > 2$ , et denominator maior est qvam $6$ .
$\delta = \min(1, 6\varepsilon)$
Hac restrictione posita qvotiens minor est qvam $\frac{\abs{x - 3}}{6}$ , et svfficit vt haec qvantitas minor sit qvam $\varepsilon$ .
$\abs{\frac{1}{x} - \frac{1}{3}} < \frac{\abs{x - 3}}{6} < \frac{6\varepsilon}{6} = \resaltar{\varepsilon}$
Si $0 < \abs{x - 3} < \delta$ , ambae condiciones implentvr. Sine restrictione praevia methodvs deficeret: prope $x = 0$ qvotiens sine margine crescit.

Problema solvtvm 4Fvnctio qvae salit

Fvnctio signi, in $\mathbb{R} \setminus \{0\}$ definita, valet $1$ in positivis et $-1$ in negativis. Demonstrare eam limitem non habere, cvm $x$ ad $0$ tendit.

s(x) = \frac{x}{\abs{x}}

Solvtio

$(\forall L)(\exists \varepsilon > 0)(\forall \delta > 0)(\exists x)\left(0 < \abs{x} < \delta \wedge \abs{s(x) - L} \geq \varepsilon\right)$
Pvnctvm $0$ est pvnctvm accvmvlationis dominii. Ostendendvm est nvllvm nvmervm realem $L$ limitem esse; per negationem definitionis svfficit pro vnoqvoqve $L$ provocationem invenire cvi nvllvm $\delta$ respondeat. Proponitvr $\varepsilon = 1$ pro omnibvs $L$ .
$s\left(\frac{\delta}{2}\right) = 1, \qquad s\left(-\frac{\delta}{2}\right) = -1$
Sit $\delta > 0$ qvodlibet. Intervallvm perforatvm radii $\delta$ nvmervm positivvm et negativvm continet, in qvibvs fvnctio valet $1$ et $-1$ .
$2 = \abs{1 - (-1)} \leq \abs{1 - L} + \abs{L - (-1)}$
Distantia inter $1$ et $-1$ est $2$ ; per inaeqvalitatem triangvlarem distantiae ipsivs $L$ ab his dvobvs valoribvs svmmam saltem $2$ efficivnt.
$\abs{s\left(\frac{\delta}{2}\right) - L} \geq 1 \vee \abs{s\left(-\frac{\delta}{2}\right) - L} \geq 1$
Inde seqvitvr alteram dvarvm distantiarvm saltem $1$ esse: si ambae minores essent qvam $1$ , svmma earvm minor esset qvam $2$ .
$\neg (\exists L)\left(\lim_{x \to 0} s(x) = L\right)$
Ita pro vnoqvoqve $L$ provocatio $\varepsilon = 1$ responsionem non habet: nvlla fascia latitvdinis $2$ simvl valores $1$ et $-1$ continet, oris exclvsis. Fvnctio in $0$ limitem non habet.

Problema solvtvm 5Maximvm delta

Pro fvnctione $f(x) = x^2$ in pvncto $a = 1$ , cvm $L = 1$ et provocatione $\varepsilon = \frac{1}{2}$ , maximvm $\delta$ determinare qvod provocationi respondet.

Solvtio

$\abs{x^2 - 1} < \frac{1}{2} \leftrightarrow \frac{1}{2} < x^2 < \frac{3}{2}$
Condicio $\abs{x^2 - 1} < \frac{1}{2}$ significat $x^2$ inter $\frac{1}{2}$ et $\frac{3}{2}$ esse.
$\sqrt{\frac{1}{2}} < x < \sqrt{\frac{3}{2}}$
Prope $1$ nvmeri $x$ positivi svnt, et condicio aeqvivalet ei qvod $x$ inter radices qvadratas terminorvm est.
$1 - \sqrt{\frac{1}{2}} \approx 0{,}293, \qquad \sqrt{\frac{3}{2}} - 1 \approx 0{,}225$
Distantiae ipsivs $1$ a terminis diversae svnt: parabola ad dextram ipsivs $1$ celerivs ascendit, et terminvs dexter propior est.
$\delta = \sqrt{\frac{3}{2}} - 1 = \resaltar{\frac{\sqrt{6} - 2}{2}}$
Intervallvm centri $1$ et radii $\delta$ in intervallo invento contineri debet; radivs eivs minorem dvarvm distantiarvm excedere non potest, qvae provocationi respondet.
$0 < \sqrt{\frac{3}{2}} - 1 < \delta', \qquad \abs{\left(\sqrt{\frac{3}{2}}\right)^2 - 1} = \frac{1}{2}$
Nvllvs radivs maior svfficit: si $\delta' > \delta$ , pvnctvm $x = \sqrt{\frac{3}{2}}$ a $1$ minvs qvam $\delta'$ distat, et in eo distantia ipsivs $x^2$ a $1$ est exacte $\frac{1}{2}$ , qvae non est minor qvam $\varepsilon$ .

Problemata proposita

Problemata difficvltate crescvnt: dvo prima, aliis polynomiis, artem radii restringendi repetvnt; tertivm negationem definitionis ad fvnctionem qvae oscillat adhibet; qvartvm examinat qvid fiat si ordo qvantificatorvm invertitvr; vltimvm fvnctionem stvdet qvae in vno tantvm pvncto limitem habet. Vnvmqvodqve indicivm habet, qvod aperire convenit solvm postqvam problema temptatvm est.

Problema propositvm 1Alivd polynomivm secvndi gradvs

Ex definitione demonstrare $\lim_{x \to -1} (2x^2 - 3x) = 5$ .

Indicivm

Resolve $2x^2 - 3x - 5$ in factores: radicem $-1$ habet. Restringe radivm ad $1$ et altervm factorem circvmscribe in intervallo qvod a $-2$ ad $0$ patet.

Responsvm

Cvm $2x^2 - 3x - 5 = (x + 1)(2x - 5)$ et, si $\abs{x + 1} < 1$ , tvm $\abs{2x - 5} < 9$ , respondet:

\delta = \min\left(1, \frac{\varepsilon}{9}\right)

Problema propositvm 2Cvbvs

Ex definitione demonstrare $\lim_{x \to 1} x^3 = 1$ .

Indicivm

Vtere factorizatione differentiae cvborvm, $x^3 - 1 = (x - 1)(x^2 + x + 1)$ , et factorem secvndvm circvmscribe, cvm $x$ inter $0$ et $2$ est.

Responsvm

Si $\abs{x - 1} < 1$ , tvm $0 < x^2 + x + 1 < 7$ , ita vt respondeat:

\delta = \min\left(1, \frac{\varepsilon}{7}\right)

Problema propositvm 3Fvnctio qvae oscillat

Demonstrare fvnctionem $\sin\left(\frac{1}{x}\right)$ , in $\mathbb{R} \setminus \{0\}$ definitam, limitem non habere, cvm $x$ ad $0$ tendit.

Indicivm

Methodvm problematis solvti 4 seqvere. In omni intervallo perforato centri $0$ svnt pvncta in qvibvs fvnctio valet $1$ et pvncta in qvibvs valet $-1$ : qvaere $x$ formae $\frac{1}{\frac{\pi}{2} + 2n\pi}$ et $\frac{1}{\frac{3\pi}{2} + 2n\pi}$ , cvm $n$ natvrali satis magno.

Responsvm

Pro vnoqvoqve $L$ nvllvm $\delta$ provocationi $\varepsilon = 1$ respondet, qvia pro $n$ magno dvo pvncta qvae seqvvntvr a $0$ minvs qvam $\delta$ distant, et in iis fvnctio valet $1$ et $-1$ :

x_n = \frac{1}{\frac{\pi}{2} + 2n\pi}, \qquad y_n = \frac{1}{\frac{3\pi}{2} + 2n\pi}

Solvtio explicata

Solvtio

$0 < \frac{1}{\frac{3\pi}{2} + 2n\pi} < \frac{1}{\frac{\pi}{2} + 2n\pi} < \frac{1}{2n\pi} < \resaltar{\delta}$
Sint $L$ nvmervs realis et $\delta > 0$ . Per proprietatem Archimedeam exsistit nvmervs natvralis $n$ cvm $2n\pi > \frac{1}{\delta}$ ; tvm dvo pvncta qvae seqvvntvr positiva svnt et minora qvam $\delta$ .
$\sin\left(\frac{\pi}{2} + 2n\pi\right) = 1, \qquad \sin\left(\frac{3\pi}{2} + 2n\pi\right) = -1$
In primo $\frac{1}{x}$ est angvlvs cvivs sinvs est $1$ ; in altero, angvlvs cvivs sinvs est $-1$ .
$\abs{1 - L} + \abs{L + 1} \geq 2 \rightarrow \abs{1 - L} \geq 1 \vee \abs{-1 - L} \geq 1$
Vt in problemate solvto 4, distantiae ipsivs $L$ ab $1$ et ab $-1$ svmmam saltem $2$ efficivnt, et altera earvm saltem $1$ est. Provocatio $\varepsilon = 1$ responsionem non habet, pro nvllo $L$ .

Problema propositvm 4Ordo qvantificatorvm

Sint $f$ fvnctio in $A$ definita, $a$ pvnctvm accvmvlationis copiae $A$ et $L$ nvmervs realis. Demonstrare envntiatvm qvod seqvitvr, qvod ex definitione limitis oritvr dvobvs primis qvantificatoribvs permvtatis, vervm esse si et solvm si $f$ valet $L$ in omnibvs pvnctis copiae $A$ alicvivs intervalli perforati centri $a$ . Satisfacitne ei fvnctio $x^2$ in $a = 3$ cvm $L = 9$ ?

(\exists \delta > 0)(\forall \varepsilon > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)

Indicivm

Si nvmervs non negativvs minor est qvam omne $\varepsilon > 0$ , est $0$ : aliter ipse esset $\varepsilon$ qvo non minor est.

Responsvm

Fixo $\delta$ envntiati, qvodqve $x$ intervalli perforati condicioni $\abs{f(x) - L} < \varepsilon$ satisfacit pro omni $\varepsilon > 0$ , ergo $f(x) = L$ ; conversa immediata est. Fvnctio $x^2$ ei in $3$ non satisfacit, qvamvis limes eivs ibi sit $9$ , qvia in nvllo intervallo constans est:

\left(3 + \frac{\delta}{2}\right)^2 \neq 9

Solvtio explicata

Solvtio

$\dato{t}{(\forall \varepsilon > 0)\left(\abs{f(x) - L} < \varepsilon\right)}$
Ponamvs envntiatvm vervm esse, et sit $\delta$ radivs qvem affirmat. Svmamvs $x$ copiae $A$ cvm $0 < \abs{x - a} < \delta$ ; inaeqvalitas pro omni $\varepsilon$ positivo valet.
$\abs{f(x) - L} < \abs{f(x) - L} \rightarrow \bot, \qquad \resaltar{f(x) = L}$
Si $\abs{f(x) - L}$ positivvm esset, svmi posset $\varepsilon = \abs{f(x) - L}$ , et inaeqvalitas diceret nvmervm se ipso minorem esse. Ergo est $0$ , et $f(x) = L$ .
$0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} = 0 < \varepsilon$
Vice versa, si $f$ valet $L$ in pvnctis copiae $A$ intervalli perforati radii $\delta$ , idem $\delta$ pro omni $\varepsilon$ svfficit, qvia distantia est $0$ .
$0 < \abs{x - 3} < 6 \rightarrow x^2 \neq 9$
Fvnctio $x^2$ non valet $9$ in vllo pvncto a $3$ diverso prope id sito, ita vt envntiato non satisfaciat, qvamvis definitioni limitis satisfaciat. Envntiatvm inversvm non est alivs modvs idem dicendi, sed assertio mvlto fortior.

Problema propositvm 5Limes in vno tantvm pvncto

Fvnctio qvae seqvitvr valet $x$ in rationalibvs et $0$ in irrationalibvs. Demonstrare limitem eivs in $0$ esse $0$ , et eam in nvllo pvncto $a \neq 0$ limitem habere.

h(x) = \begin{cases} x & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}

Indicivm

In $0$ observa $\abs{h(x)} \leq \abs{x}$ in omnibvs casibvs. In $a \neq 0$ propone provocationem $\varepsilon = \frac{\abs{a}}{3}$ et vtere eo qvod in omni intervallo svnt rationales et irrationales, vt in lectione de recta reali demonstratvm est: irrationalis propinqvvs cogit $L$ prope $0$ esse, et rationalis propinqvvs, prope $a$ .

Responsvm

In $0$ respondet $\delta = \varepsilon$ . In $a \neq 0$ , si $L$ limes esset, irrationalis $y$ et rationalis $x$ intervalli perforati radii $\min\left(\delta, \frac{\abs{a}}{3}\right)$ simvl darent dvas inaeqvalitates qvae seqvvntvr, qvae inter se repvgnant:

\abs{L} < \frac{\abs{a}}{3}, \qquad \abs{L} > \abs{x} - \frac{\abs{a}}{3} > \frac{\abs{a}}{3}

Solvtio explicata

Solvtio

$0 < \abs{x} < \varepsilon \rightarrow \abs{h(x) - 0} \leq \abs{x} < \varepsilon$
In $0$ : in rationalibvs $\abs{h(x)} = \abs{x}$ , et in irrationalibvs $\abs{h(x)} = 0$ ; vtroqve in casv distantia ipsivs $h(x)$ ab $0$ non excedit distantiam ipsivs $x$ ab $0$ , et respondet $\delta = \varepsilon$ .
$\delta' = \min\left(\delta, \frac{\abs{a}}{3}\right)$
Sit $a \neq 0$ et ponamvs $L$ esse limitem in $a$ . Cvm $\varepsilon = \frac{\abs{a}}{3}$ est $\delta$ qvod respondet, idqve, si opvs est, ad $\frac{\abs{a}}{3}$ minvitvr.
$\abs{0 - L} < \frac{\abs{a}}{3} \rightarrow \dato{i}{\abs{L} < \frac{\abs{a}}{3}}$
In intervallo perforato radii $\delta'$ est irrationalis $y$ , in qvo $h$ valet $0$ : limes prope $0$ manet.
$\abs{x - L} < \frac{\abs{a}}{3}, \qquad \dato{r}{\abs{x} > \abs{a} - \frac{\abs{a}}{3} = \frac{2\abs{a}}{3}}$
Est etiam rationalis $x$ , in qvo $h$ valet $x$ . Cvm $x$ ab $a$ minvs qvam $\frac{\abs{a}}{3}$ distet, valor eivs absolvtvs maior est qvam $\frac{2\abs{a}}{3}$ .
$\abs{L} \geq \abs{x} - \abs{x - L} > \frac{2\abs{a}}{3} - \frac{\abs{a}}{3} = \resaltar{\frac{\abs{a}}{3}}$
Per inaeqvalitatem triangvlarem $\abs{L}$ simvl minor et maior esset qvam $\frac{\abs{a}}{3}$ : contradictio. In $a$ limes non est.