Limit depth: epsilon planning, domains and failed candidates

Learning goals and definitions

Build a delta from a target epsilon, control a denominator before division, and use domain and sequence counterexamples to disprove a limit.

Let $A\subseteq\mathbb R$ and let $a$ be an accumulation point of $A$: every punctured neighborhood contains a point of $A$. The relative-domain limit $f(x)\to L$ as $x\to a$, $x\in A$, means that for every $\varepsilon>0$ there is $\delta>0$ such that $x\in A$ and $0<|x-a|<\delta$ imply $|f(x)-L|<\varepsilon$. The delta may depend on epsilon, the fixed anchor and the given function, but not on the eventual approaching input.

The puncture excludes $a$ itself. A limit can exist even if $f(a)$ is absent or different from $L$. Continuity at $a$ additionally requires $a\in A$ and $f(a)=L$ with the appropriate relative approach. A one-sided limit further restricts the approaching points to $x>a$ or $x<a$. The domain must contain points arbitrarily close to $a$ on the specified side.

Proposition: uniqueness needs an accumulation point

A finite relative-domain limit at an accumulation point is unique.

Proof. If different limits $L,M$ existed, take $\varepsilon=|L-M|/3$ and the smaller of their two deltas. Accumulation supplies an actual domain point in that punctured neighborhood. Both limit inequalities hold there, so $|L-M|\le|L-f(x)|+|f(x)-M|<2|L-M|/3$, impossible. Without an approaching domain point, the quantified implication could be vacuously true for every proposed value.

1. Plan backward, then write the proof forward

To prove $x^2\to a^2$, factor the target error as $|x-a||x+a|$. The second factor still varies. First require $|x-a|<1$, giving $|x+a|<1+2|a|$. Then the sufficient condition $|x-a|<\varepsilon/(1+2|a|)$ is available. Choose

$$\delta=\min\left(1,\frac{\varepsilon}{1+2|a|}\right).$$

For every input within that delta, the product error is below epsilon. The proof works at $a=0$ as well as elsewhere because its denominator never vanishes. The extra condition $|x-a|<1$ is a buffer bound, not an arbitrary loss of generality.

For $f(x)=1/x$ at $a\ne0$, the error is $|x-a|/(|x||a|)$. First impose $|x-a|<|a|/2$, so $|x|>|a|/2$ and the denominator stays away from zero. Now

$$\left|\frac1x-\frac1a\right|<\frac{2|x-a|}{a^2}, \qquad \delta=\min\left(\frac{|a|}{2},\frac{\varepsilon a^2}{2}\right)$$

proves the result. An inequality using an uncontrolled $1/|x|$ would not yet justify a delta depending only on epsilon and the fixed anchor.

The planning can be written backward on scratch paper. The finished proof must run forward: fix arbitrary epsilon, specify delta, take every allowed input within it, and derive the target bound. Checking one selected input after choosing delta does not establish the universal statement.

2. Domains determine which sides exist

For $f(x)=|x|/x$ on nonzero reals, the right-hand limit at zero is one and the left-hand limit minus one, so no finite two-sided limit exists. Restricting the domain to positive inputs changes the relative limit to one. A limit is a property of a function with a domain, not only of its written formula.

For $f(x)=\sqrt x$ on $[0,\infty)$, the relative limit at zero is zero, proved by $0\le x<\varepsilon^2\Rightarrow\sqrt x<\varepsilon$. The domain does not approach zero from the left. The theorem equating a two-sided limit with equal one-sided limits applies when the domain approaches from both sides; an unavailable side is not an additional required condition in this boundary example.

For $(x^2-1)/(x-1)$ on $x\ne1$, factorization gives limit two at one. Assigning the missing value two makes a continuous extension; assigning five gives a discontinuous function with the same punctured limit. The isolated point value cannot repair a genuinely oscillating punctured neighborhood.

3. Bounding oscillation and proving failure are different methods

On nonzero inputs, $|x\sin(1/x)|\le|x|$, so the squeeze estimate proves its limit zero. The oscillating factor itself has no limit, but boundedness of that factor is enough when multiplied by something tending to zero.

To prove $\sin(1/x)$ has no limit even from the right, choose $x_n=1/(\pi/2+2n\pi)$ and $y_n=1/(3\pi/2+2n\pi)$. Both approach zero through positive inputs, but their function values are one and minus one. Any finite limit would force values along both sequences to approach it, impossible.

The sequence principle used here follows directly from the definition: if a limit exists, every approaching sequence of allowed punctured-domain points eventually lies within its delta and hence has the same output limit. Conversely, if the limit definition fails, some fixed $\varepsilon_0>0$ permits a bad point within distance $1/n$ for every $n$; selecting those points gives an approaching sequence whose errors never drop below $\varepsilon_0$. Thus testing every sequence is an equivalent criterion. Testing two convenient sequences is enough to disprove a limit when they disagree, but their agreement is not enough to prove it.

4. Infinite endpoints and infinite values use different thresholds

The finite limit $f(x)\to L$ as $x\to+\infty$ means every epsilon has an input threshold $M$ beyond which the output error is below epsilon. At negative infinity, approaching inputs satisfy $x<M$ for a sufficiently negative threshold. For an infinite-endpoint limit, the domain must be unbounded in the corresponding direction. Neither infinity is a real input at which to substitute.

The statement $f(x)\to+\infty$ as $x\to a$ instead means that every positive height $B$ is eventually exceeded in a punctured neighborhood. For $1/x^2$ at zero, choosing $\delta=1/\sqrt B$ proves $0<|x|<\delta\Rightarrow1/x^2>B$. This is not a finite limit. For $1/x$, opposite sides give opposite infinite signs; only its absolute magnitude tends to positive infinity two-sidedly.

Worked tail estimate. Let $f(x)=(2x^2+x)/(x^2+1)$. Its difference from two is $(x-2)/(x^2+1)$. For $|x|\ge1$, its magnitude is at most $1/|x|+2/x^2\le3/|x|$. A positive threshold above $\max(1,3/\varepsilon)$ proves the limit two at positive infinity; requiring $x$ below the negative of that threshold proves the same at negative infinity. The estimates, not a substitution of infinity, establish both directions.

Graded proof and diagnostic tasks

1. Boundary versus interior. Give explicit delta choices for $\sqrt x\to\sqrt a$ at $a>0$ on its real domain and at $a=0$. Explain why the two choices naturally have different forms.
2. Buffer audit. A student proposes $\delta=\varepsilon a^2$ for the reciprocal limit without controlling $|x|$. Explain why this choice need not work and give a corrected general choice.
3. Domain restriction. On $A=\{1/n:n\ge1\}\cup\{-1/n:n\ge1\}$ define $f(x)$ to be the sign of $x$. Determine its relative limit at zero, then repeat after restricting to the positive part of $A$.
4. Accumulation condition. For $A=\{0\}\cup[1,\infty)$, show why the punctured epsilon-delta implication at zero becomes true for every proposed $L$ if the accumulation requirement is omitted.
5. Two tails. Use the rational-tail estimate above to write a complete proof of both infinity-direction limits, explicitly naming the positive threshold and its negative counterpart.
6. Method choice. Prove $x^2\sin(1/x)\to0$ without a limit for the sine factor. What condition on a multiplier $b(x)$ is sufficient for $x^2b(x)\to0$, and is it necessary?

Full solutions

1. For $a>0$ and $x\ge0$, rationalization gives $|\sqrt x-\sqrt a|=|x-a|/(\sqrt x+\sqrt a)\le|x-a|/\sqrt a$. Thus $\delta=\varepsilon\sqrt a$ suffices on the relative domain. At zero use $\delta=\varepsilon^2$, since $0\le x<\delta$ implies $\sqrt x<\varepsilon$. The positive anchor provides a fixed denominator in the first proof, absent in the second.
2. At $a=1,\varepsilon=1$, that proposal gives delta one; the permitted input $x=1/10$ has distance $9/10<1$ but reciprocal error nine, not below one. A valid choice is $\min(|a|/2,\varepsilon a^2/2)$, which first bounds the denominator away from zero and then controls the numerator.
3. The sequences $1/n$ and $-1/n$ both approach zero but have constant output values one and minus one, so no relative finite limit exists on all of $A$. On the positive restricted domain the function is identically one, hence has relative limit one. The domain change removes the competing approach, not the algebraic formula's sign distinction.
4. Choose any delta below one. There is no domain point with $0<|x|<\delta$. Therefore the implication places no condition on outputs and is true for every epsilon and every proposed $L$. Zero is not an accumulation point of that domain, so the standard limit definition excludes this vacuity and preserves uniqueness.
5. Given epsilon, choose $M>\max(1,3/\varepsilon)$. For $x>M$, the derived error is at most $3/|x|<3/M<\varepsilon$. For $x<-M$, the same magnitude bound holds because $|x|>M$. These are separate positive- and negative-tail proofs using a shared estimate.
6. The sine magnitude is at most one, so the product magnitude is at most $x^2$, tending to zero; delta $\sqrt\varepsilon$ suffices. Any multiplier bounded in some punctured neighborhood supplies a constant $C$ and bound $C x^2\to0$. Boundedness is sufficient but not necessary: $b(x)=1/|x|$ is unbounded yet $x^2b(x)=|x|\to0$.

Summary and terms

A buffer bound controls a varying factor before choosing a final error threshold. A relative-domain limit uses only actual approaching domain points. An accumulation point prevents a vacuous punctured implication. A counterexample sequence can refute a proposed limit but finitely many agreeing paths cannot certify it. Keep input thresholds, output tolerances and infinite-height thresholds distinct in the proof.

Proofs supporting the methods used in this note

Finite real limit laws and geometric sequences: the missing proofs

Work over the ordered real numbers, with the Archimedean property, positive-integer induction and the usual absolute-value triangle inequality. A sequence converges to a finite real $L$ when every $\varepsilon>0$ admits an $N$ that works for every $n\ge N$.

Uniqueness, boundedness and subsequences. If $a_n\to L,M$ and $L\ne M$, require both errors below $|L-M|/3$ at the same sufficiently large index. The triangle inequality gives $|L-M|<2|L-M|/3$, a contradiction. For boundedness, require $|a_n-L|<1$ in the tail, and take a maximum of $|L|+1$ and the finitely many initial absolute values. For a subsequence $n_1<n_2<\cdots$, induction gives $n_k\ge k$; hence every tail threshold for the original sequence also works eventually for the subsequence.

Sum, scalar and product. If $a_n\to a$, $b_n\to b$, choose both input errors below $\varepsilon/2$; then

$$|(a_n+b_n)-(a+b)|\le |a_n-a|+|b_n-b|<\varepsilon.$$

For a fixed scalar $c\ne0$ use input tolerance $\varepsilon/|c|$; $c=0$ is immediate. Choose $B\ge1$ bounding all $|b_n|$. Require

$$|a_n-a|<\frac{\varepsilon}{2B},\qquad |b_n-b|<\frac{\varepsilon}{2(|a|+1)}.$$

The common later threshold gives $|a_nb_n-ab|\le B|a_n-a|+|a||b_n-b|<\varepsilon$.

Reciprocal and quotient. If $b\ne0$, first require $|b_n-b|<|b|/2$, ensuring $|b_n|>|b|/2$ and nonzero denominators. Then

$$\left|\frac1{b_n}-\frac1b\right|\le\frac{2|b_n-b|}{|b|^2}.$$

Requiring also $|b_n-b|<\varepsilon|b|^2/2$ proves reciprocal convergence. The product law now proves $a_n/b_n\to a/b$. Any undefined initial quotients may be omitted; convergence is a tail assertion.

Order and squeeze. If $a_n\le b_n$ eventually and $a_n\to a$, $b_n\to b$, then $a\le b$: otherwise errors below $(a-b)/3$ would force $a_n>b_n$. If $a_n\le z_n\le b_n$ eventually and both outer limits are $L$, require both errors below $\varepsilon$. Then $L-\varepsilon<z_n<L+\varepsilon$, proving $z_n\to L$. This proof does not require a previously known limit of $z_n$.

For the corresponding function-limit laws, fix a common allowed approach domain $D\subseteq\mathbb R$ and an accumulation point $a$ of $D$: every punctured neighborhood of $a$ contains a point of $D$. Let $f,g:D\to\mathbb R$ have the stated finite relative-domain limits at $a$. Replace each tail threshold in the displayed estimates by a punctured-domain radius, and take the minimum of finitely many positive radii. For the product proof only local boundedness is needed: $g(x)\to b$ gives $|g(x)|\le|b|+1$ on a sufficiently small punctured $D$-neighborhood, even if $g$ is unbounded elsewhere on $D$. The same estimates prove sum, scalar, product, reciprocal and quotient laws with their stated nonzero limiting-denominator condition; the same order and squeeze estimates apply on the common approach domain. Accumulation supplies actual points when proving uniqueness or excluding an incompatible order of limiting values.

For composition let $B\subseteq\mathbb R$, let $b\in B$, and let $F:B\to\mathbb R$ be continuous at $b$ relative to $B$. Suppose $g:D\to\mathbb R$ tends to $b$ and $g(x)\in B$ throughout some sufficiently small punctured $D$-neighborhood of $a$. Given $\varepsilon>0$, continuity supplies $\eta>0$ such that $u\in B$ and $|u-b|<\eta$ imply $|F(u)-F(b)|<\varepsilon$. The limit of $g$ and the composition-domain condition supply a common positive radius forcing both conditions for every allowed approaching $x$. Thus $F(g(x))\to F(b)$. This includes inputs with $g(x)=b$ and does not divide by an inner increment.

Geometric classification. Bernoulli's bound $(1+t)^n\ge1+nt$ for $t\ge0$ follows by induction: multiply the case $n$ by the positive $1+t$ and retain the nonnegative remainder $nt^2$. If $0<q<1$, set $t=q^{-1}-1>0$. Thus

$$0<q^n\le\frac1{1+nt}\longrightarrow0.$$

For any requested tolerance, the Archimedean property supplies $N$ making the right side smaller. Hence $r^n\to0$ for $0<|r|<1$ by $|r^n|=|r|^n$; $r=0$ gives zero for every positive index. At $r=1$ the sequence is constantly one. At $r=-1$, the even and odd subsequences are one and minus one, so uniqueness and subsequence inheritance exclude convergence. If $r>1$, Bernoulli with $t=r-1$ gives $r^n\ge1+nt\to+\infty$. If $r<-1$, $|r^n|\to+\infty$ by the same argument, contradicting boundedness of any convergent real sequence. These cases exhaust all real ratios.