2.2 Sequences and multivariable limits

A sequence approaches a point by choosing one input after another. A function limit makes a stronger-looking demand: every sufficiently nearby admissible input must give a nearby output. The sequential criterion will show that these descriptions express the same mathematical control. It also explains the different jobs of the two main techniques for multivariable limits. A carefully chosen path can disprove a candidate limit; an estimate valid for all nearby points can prove one.

Keep the domain visible throughout. A function may be defined on a punctured plane, a half-plane, or a curve. A limit is taken through points of that domain. We do not evaluate the function along a deleted line, and we do not need to assign a value at the point being approached. What matters is whether admissible points occur arbitrarily close to that point.

Convergence of vector sequences

Definition

Convergence and subsequences

A sequence $x_k\in\mathbb R^n$ converges to $a$ if, for every $\varepsilon>0$, there is an integer $N$ such that

$$k\ge N\quad\Longrightarrow\quad\|x_k-a\|<\varepsilon.$$

A subsequence has the form $x_{k_j}$, where $k_1<k_2<\cdots$ are positive integer indices.

The entire tail, rather than selected terms, must enter each ball. This is why a bounded sequence can fail to converge: it may keep visiting two separated regions. The sequence $(({-1})^k,1/k)$ has an even-index subsequence approaching $(1,0)$ and an odd-index subsequence approaching $(-1,0)$. Since the limits differ, the original sequence does not converge.

Theorem

Coordinate convergence

A sequence in $\mathbb R^n$ converges to $a$ exactly when every coordinate converges to the corresponding coordinate of $a$.

Proof

One distance controls all coordinates

For each coordinate,

$$|x_{k,i}-a_i|\le\|x_k-a\|.$$

Thus norm convergence implies coordinate convergence. Conversely, given $\varepsilon>0$, coordinate convergence supplies an index after which each coordinate error is less than $\varepsilon/\sqrt n$. There are finitely many coordinates, so take the largest of these indices. For every later term,

$$\|x_k-a\|^2=\sum_{i=1}^n|x_{k,i}-a_i|^2<\varepsilon^2.$$

The finite number of coordinates is what allows a common index to be chosen.

For example, $(2/k,\cos(1/k),k/(k+3))$ converges to $(0,1,1)$. Computing coordinates is efficient, but the norm definition remains the statement that makes convergence independent of how we write the coordinates.

What a function limit requires

Definition

A limit relative to a domain

Let $F:D\subseteq\mathbb R^n\to\mathbb R^m$, and let $a$ be an accumulation point of $D$. We write $\lim_{x\to a,\ x\in D}F(x)=L$ if

$$\forall\varepsilon>0\ \exists\delta>0\ \forall x\in D: \quad 0<\|x-a\|<\delta\ \Longrightarrow\ \|F(x)-L\|<\varepsilon.$$

The accumulation-point hypothesis guarantees that punctured balls contain admissible inputs. Without it, the implication could be true simply because there are no such inputs. For a scalar output, the output norm is absolute value. The condition excludes the centre, so changing $F(a)$ cannot change this limit.

The order of the quantifiers gives a practical instruction. First choose an output tolerance. Then choose one input radius that works for every admissible point inside the punctured ball. The radius may depend on the tolerance and on the fixed target point, but it cannot be adjusted after a path or input point has been selected.

Limits are unique. If two distinct values $L,M$ were limits, use the output tolerance $\|L-M\|/3$ for both. A sufficiently close admissible input would then give

$$\|L-M\|\le\|L-F(x)\|+\|F(x)-M\|<2\|L-M\|/3,$$

a contradiction. Notice where the accumulation-point hypothesis is used: it supplies an input inside the common punctured ball.

Theorem

Sequential criterion

The limit $F(x)\to L$ as $x\to a$ through $D$ holds exactly when, for every sequence $x_k\in D\setminus\{a\}$ with $x_k\to a$, the output sequence satisfies $F(x_k)\to L$.

Proof

Turning failure of uniform control into a sequence

If the function limit holds, choose its radius $\delta$ for a given $\varepsilon$. The input sequence eventually lies within that punctured radius, so the output sequence eventually lies within $\varepsilon$ of $L$.

Conversely, suppose the function limit fails for $L$. There is then a fixed $\varepsilon_0>0$ such that every positive radius contains an admissible point with output error at least $\varepsilon_0$. For radius $1/k$, choose one such $x_k$. Then

$$0<\|x_k-a\|<1/k,\qquad \|F(x_k)-L\|\ge\varepsilon_0.$$

The input sequence converges to $a$, but its output sequence cannot converge to $L$. This contradicts the sequential condition.

Proving a limit with estimates

The algebraic laws remain valid in several variables. Sums, constant multiples and products of scalar functions with finite limits have the expected limits. The quotient law requires the denominator limit to be nonzero. One useful reason these proofs work is local boundedness: if $f(x)\to L$, then sufficiently close inputs satisfy $|f(x)|<|L|+1$.

For products, write

$$fg-LM=f(g-M)+M(f-L).$$

Bound the varying factor $f$ first, then make the two remaining errors small. This pattern is more useful than memorising a separate proof for every polynomial expression.

Worked example

An explicit epsilon–delta choice at a nonzero point

Prove that $xy\to-2$ as $(x,y)\to(-1,2)$. Put $d=\|(x,y)-(-1,2)\|$. The error satisfies

$$|xy+2| =|x(y-2)+2(x+1)| \le |x|\,|y-2|+2|x+1|.$$

If $d<1$, then $|x|<2$. Each coordinate error is at most $d$, so $|xy+2|<4d$. Given $\varepsilon>0$, take

$$\delta=\min\{1,\varepsilon/4\}.$$

When $0<d<\delta$, the preliminary local bound and the final error estimate both hold. Therefore $|xy+2|<\varepsilon$.

The first bound did not attempt to make the output error small; it made a varying coefficient controllable. The second part of the radius then imposed the desired tolerance. This two-stage choice is typical for nonlinear expressions.

Theorem

A uniform bound proves a zero limit

Suppose that, for all sufficiently close admissible $x\ne a$,

$$\|F(x)\|\le C\|x-a\|^\alpha$$

with constants $C>0$ and $\alpha>0$. Then $F(x)\to0$. More generally, a scalar function trapped between two functions with the same limit has that limit.

For the displayed bound, choose $\delta$ no larger than the radius where the bound is valid and no larger than $(\varepsilon/C)^{1/\alpha}$. This is a precise form of the squeeze method: one distance estimate controls all directions at once.

Disproving a limit by paths

A path $\gamma(t)\to a$ whose points lie in $D\setminus\{a\}$ supplies admissible inputs for a punctured limit. The sequential criterion forces its output to approach the same limit whenever the multivariable limit exists. Two such paths with different output limits therefore disprove existence. The paths need not be straight.

Worked example

All straight lines can miss the obstruction

On the punctured plane, let

$$f(x,y)=\frac{x^3y}{x^6+y^2}.$$

Along $y=mx$ with $m\ne0$,

$$f(x,mx)=\frac{m x^2}{x^4+m^2}\longrightarrow0.$$

The horizontal and vertical axes also give zero. Thus every straight line through the origin gives the same limit.

However, along the curve $y=x^3$, the value is $1/2$ for every $x\ne0$. Both this curve and the horizontal axis stay in the domain away from the origin, and both approach the origin. Their output limits differ, so the multivariable limit does not exist.

The denominator suggests the curve: balance $x^6$ against $y^2$, giving $y$ of order $x^3$. A line usually makes one denominator term dominate, hiding the balance that produces the obstruction.

This argument does not say that curves are always needed. Simple axes or lines often settle the question. It says that agreement on a convenient collection of paths supplies evidence for a conjecture, while a proof of existence needs control over every nearby domain point.

Iterated limits are another restricted test. For the function in the example, taking one variable to zero with the other fixed gives zero in either order, so both iterated limits equal zero. They still do not describe simultaneous approach along $y=x^3$. Equal iterated limits alone do not establish a multivariable limit.

Uniformity in polar coordinates and weighted bounds

Writing $x=r\cos\theta$, $y=r\sin\theta$ changes approach to the origin into $r\to0$, while $\theta$ may vary with $r$. A valid estimate such as $|f(r\cos\theta,r\sin\theta)|\le Cr^\alpha$ needs one constant $C$ for all relevant angles. A statement about each fixed angle does not provide this uniform control.

For instance, $x^2y/(x^2+y^2)=r\cos^2\theta\sin\theta$ has absolute value at most $r$. The angular factor is uniformly bounded, so the limit is zero. The estimate, rather than the substitution itself, completes the proof.

Worked example

An extra power restores a limit

Consider

$$g(x,y)=\frac{x^4y}{x^6+y^2},\qquad(x,y)\ne(0,0).$$

The inequality $(|x|^3-|y|)^2\ge0$ gives

$$2|x^3y|\le x^6+y^2.$$

Consequently

$$|g(x,y)|=|x|\frac{|x^3y|}{x^6+y^2}\le\frac{|x|}{2} \le\frac{\sqrt{x^2+y^2}}2.$$

This bound holds at every domain point and tends to zero, so $g\to0$. In epsilon–delta form, $\delta=2\varepsilon$ works.

Compare this with the preceding example: the problematic ratio is still present, but the additional factor $x$ makes its entire bounded contribution disappear uniformly. The argument also shows why comparing the total polynomial degrees alone is unreliable when the denominator scales differently in different variables.

Common mistake

A radius that depends on the angle

After a polar substitution, choosing a different input radius for each fixed angle proves only separate radial limits. In a multivariable limit the angle can change as the radius changes. Seek a single angular bound, or return to an inequality in $x$ and $y$ that is valid everywhere nearby.

Vector outputs and behaviour at infinity

For $F=(F_1,\ldots,F_m)$, a finite limit exists exactly when all component limits exist. The proof repeats the coordinate-convergence inequalities, using an error budget of $\varepsilon/\sqrt m$ for each component. For example,

$$F(x,y)=\left(\frac{\sin(x^2+y^2)}{x^2+y^2},\ \frac{x^4y}{x^6+y^2}\right)$$

tends to $(1,0)$ on the punctured plane, by the one-variable sine limit and the bound just proved.

At infinity the input condition becomes $\|x\|>R$ rather than $0<\|x-a\|<\delta$. One radius must again work in every direction. For

$$h(x,y)=\frac{2x-y}{1+x^2+y^2},$$

Cauchy–Schwarz gives $|h|\le\sqrt5\,r/(1+r^2)\le\sqrt5/r$ for $r>0$. Given $\varepsilon>0$, taking $R=\sqrt5/\varepsilon$ proves that $h\to0$ as $r\to\infty$.

An infinite output limit uses a different tolerance: $f(x)\to+\infty$ means that for every $M>0$, all sufficiently close admissible inputs satisfy $f(x)>M$. Thus $1/(x^2+y^2)\to+\infty$ at the origin, since choosing $\delta=1/\sqrt M$ forces the desired inequality. Infinity here describes eventual growth, rather than a real number to substitute into an expression.

How to choose and present an argument

Start by stating what is being claimed. To prove a particular candidate $L$, it is enough to control the distance from the output to that candidate. To prove that no finite limit exists, it is often cleaner to find two incompatible output behaviours. The first task has a universal input requirement; the second can be settled by particular counterexamples. Keeping these logical roles separate makes the final conclusion much easier to assess.

The negation of a proposed limit deserves careful reading. It does not say that every nearby input is bad. It says that one fixed positive output error can be exceeded by at least one admissible input in every punctured ball. The bad input may depend on the radius. Choosing one bad input for each reciprocal radius produces the sequence in the sequential-criterion proof. Thus even a very thin family of bad inputs can destroy a limit, while a large collection of good inputs cannot by itself establish one.

A useful preliminary result is that finite limits force local boundedness. Choose output tolerance one in the definition. Eventually the output lies within distance one of its finite target, so the triangle inequality bounds its norm by the target norm plus one. This is the same reason a convergent sequence is bounded: control its tail and then enlarge the bound to include the finitely many earlier terms. A function need not be bounded on its whole domain to be bounded close to the point of approach.

The denominator condition in a quotient law has a similar geometric meaning. If a denominator tends to a nonzero number, choose a tolerance less than half the magnitude of that number. Nearby denominator values then stay a positive distance from zero, so their reciprocals are uniformly bounded. The quotient error can consequently be estimated using the numerator and denominator errors. If the denominator tends to zero, this argument loses its bound, and a separate analysis is needed. An expression producing an indeterminate form has reached precisely this missing step.

When using a path, write the path as an actual input map and check its domain. One may take any parameter sequence tending to its endpoint; the images of those parameters form an admissible input sequence. The function-limit sequential criterion then forces the corresponding output sequence to have the proposed limit. This is why a genuine multivariable limit controls every admissible path, even one whose direction changes infinitely often near the endpoint.

Finally, a good proof distinguishes discovery from verification. During discovery, trying lines or balancing denominator terms is an efficient way to guess the answer. In the written verification, show the concrete reason that proves the claim: two computed and incompatible limits, or one inequality valid for all nearby inputs. A long list of successful numerical trials does not supply the universal bound. Conversely, once a short uniform estimate has been found, there is no need to supplement it with further path tests.

Exercises and solutions

Checkpoint

Exercise 1: Build an explicit input radius.

Use the definition to prove $2x-3y\to5$ as $(x,y)\to(1,-1)$.

Solution · Exercise 1 solution

With $d=\|(x,y)-(1,-1)\|$, Cauchy–Schwarz gives $|2(x-1)-3(y+1)|\le\sqrt{13}\,d$. Therefore $\delta=\varepsilon/\sqrt{13}$ works. A less sharp but valid choice is $\varepsilon/5$, using the sum of coordinate bounds.

Checkpoint

Exercise 2: Detect a curved-path obstruction.

Does $x^2y/(x^4+4y^2)$ have a limit at the origin on the punctured plane?

Solution · Exercise 2 solution

Along $y=0$ the value is zero. Along $y=x^2/2$, the denominator is $2x^4$ and the numerator is $x^4/2$, so the value is $1/4$. Both paths are admissible for $x\ne0$; the different limits disprove existence.

Checkpoint

Exercise 3: Prove a three-variable limit uniformly.

Show that $xyz/(x^2+y^2+z^2)\to0$ as $(x,y,z)\to(0,0,0)$ through nonzero points.

Solution · Exercise 3 solution

Put $r=\sqrt{x^2+y^2+z^2}$. Each coordinate has absolute value at most $r$, so $|xyz|\le r^3$. Dividing by $r^2$ gives an output bound $r$, and $\delta=\varepsilon$ proves the limit. No selection of individual paths is needed.

For a new problem, identify the domain and the point of approach, try simple paths to discover a plausible result, and then choose the appropriate proof. Different path values settle nonexistence; an estimate in the distance settles existence. The sequential criterion explains why these two forms of reasoning are consistent.