The Sequential Criterion for Multivariable Limits

Why Sequences Help Test a Limit

A multivariable limit describes what happens to \(f(\mathbf{x})\) as the input \(\mathbf{x}\) approaches a point \(\mathbf{a}\) through its domain. The input can approach from infinitely many directions and along many kinds of curves. The epsilon-delta definition captures all those approaches at once, but it can be difficult to use directly.

The sequential criterion gives an equivalent way to reason: instead of checking every nearby input at once, consider every sequence of domain points that approaches \(\mathbf{a}\). If the function values always approach the same number \(L\), then the limit is \(L\). If two such sequences produce incompatible function values, the limit does not exist.

Precise Statement

Let \(D\) be a subset of \(\mathbb{R}^n\), let \(f:D\to\mathbb{R}\), and let \(\mathbf{a}\) be a limit point of \(D\). Being a limit point means that every neighborhood of \(\mathbf{a}\) contains a point of \(D\) other than \(\mathbf{a}\). The function need not be defined at \(\mathbf{a}\).

Definition: The limit \(\lim_{\mathbf{x}\to\mathbf{a}}f(\mathbf{x})=L\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that, for every \(\mathbf{x}\in D\), if \(0<\|\mathbf{x}-\mathbf{a}\|<\delta\), then \(|f(\mathbf{x})-L|<\varepsilon\).

The puncture \(0<\|\mathbf{x}-\mathbf{a}\|\) matters: the definition concerns nearby inputs other than the point itself. The norm measures distance in the input space; in \(\mathbb{R}^2\), the usual choice is \(\|(x,y)\|=\sqrt{x^2+y^2}\).

Theorem: Sequential Criterion for Multivariable Limits. With \(D\), \(f\), and \(\mathbf{a}\) as above, \(\lim_{\mathbf{x}\to\mathbf{a}}f(\mathbf{x})=L\) if and only if, for every sequence \((\mathbf{x}_k)\) with \(\mathbf{x}_k\in D\setminus\{\mathbf{a}\}\) and \(\mathbf{x}_k\to\mathbf{a}\), the sequence \(f(\mathbf{x}_k)\) converges to \(L\).

Both parts of this statement are essential. A limit forces the same output limit along every approaching sequence. Conversely, if the multivariable limit fails, at least one approaching sequence must witness that failure. In particular, if two approaching sequences yield function values with different limits, there can be no multivariable limit.

Why the Criterion Is True

The proof follows directly from the epsilon-delta definition. First, suppose the limit is \(L\), and take any sequence in the theorem. Given \(\varepsilon>0\), the definition supplies a \(\delta>0\) for which inputs within \(\delta\) of \(\mathbf{a}\), but not equal to it, satisfy \(|f(\mathbf{x})-L|<\varepsilon\). Since \(\mathbf{x}_k\to\mathbf{a}\), there is an index \(K\) such that \(k\ge K\) implies \(\|\mathbf{x}_k-\mathbf{a}\|<\delta\). The sequence terms are also not equal to \(\mathbf{a}\), so

k\ge K \quad\Longrightarrow\quad |f(\mathbf{x}_k)-L|<\varepsilon.

This is exactly the definition that \(f(\mathbf{x}_k)\to L\).

For the converse, suppose every sequence in the theorem has function values converging to \(L\), but the limit is not \(L\). Negating the epsilon-delta definition gives some \(\varepsilon_0>0\) such that for every \(\delta>0\), there is a point \(\mathbf{x}\in D\) satisfying

0<\|\mathbf{x}-\mathbf{a}\|<\delta \qquad\text{and}\qquad |f(\mathbf{x})-L|\ge\varepsilon_0.

For each \(k\in\mathbb{N}\), apply this statement with \(\delta=1/k\) and choose one such point \(\mathbf{x}_k\). Then \(0<\|\mathbf{x}_k-\mathbf{a}\|<1/k\), which implies \(\mathbf{x}_k\to\mathbf{a}\). Yet every function value remains at least \(\varepsilon_0\) from \(L\). Thus \(f(\mathbf{x}_k)\) does not converge to \(L\), contradicting the assumed behavior of every approaching sequence. The limit must therefore be \(L\).

Using Incompatible Sequences to Show Nonexistence

To disprove a proposed limit, it is enough to find two sequences in the domain that approach the same point but whose function values have different limits. The sequence terms must belong to the domain, must not equal the point being approached, and must actually converge to that point.

Worked Example: Two Straight-Line Sequences Give Different Limits

Consider \(f(x,y)=\frac{xy}{x^2+y^2}\), defined for \((x,y)\ne(0,0)\). Test whether it has a limit at the origin using two sequences.

First take \(\mathbf{x}_k=(1/k,1/k)\), for \(k\in\mathbb{N}\). Its distance from the origin is \(\sqrt{2}/k\), so \(\mathbf{x}_k\to(0,0)\), and every term lies in the domain. Substitution gives

f(1/k,1/k) = \frac{(1/k)(1/k)}{(1/k)^2+(1/k)^2} = \frac{1/k^2}{2/k^2} = \frac12.

Now take \(\mathbf{y}_k=(1/k,-1/k)\). Its distance is also \(\sqrt{2}/k\), so it too approaches the origin through the domain. This time,

f(1/k,-1/k) = \frac{(1/k)(-1/k)}{(1/k)^2+(-1/k)^2} = \frac{-1/k^2}{2/k^2} = -\frac12.

The first sequence of function values is constantly \(1/2\), and the second is constantly \(-1/2\). Their limits differ, so the sequential criterion proves that \(f\) has no limit at the origin.

Sequences need not follow straight lines. A curve can be encoded by choosing its points at a sequence of parameter values. This is useful when a formula behaves differently along a curved approach.

Worked Example: Curved Sequences Give Incompatible Values

Consider \(g(x,y)=\frac{x^2y}{x^4+y^2}\), again defined away from the origin. Choose the sequences \(\mathbf{x}_k=(1/k,1/k^2)\) and \(\mathbf{y}_k=(1/k,-1/k^2)\).

For either sequence, the squared distance to the origin is \(1/k^2+1/k^4\), which tends to zero. Each sequence therefore approaches the origin, and neither contains the origin. For the first sequence, the numerator is \((1/k)^2(1/k^2)=1/k^4\), while the denominator is \((1/k)^4+(1/k^2)^2=2/k^4\). Therefore,

g(1/k,1/k^2) = \frac{1/k^4}{2/k^4} = \frac12.

For the second sequence, the numerator is \((1/k)^2(-1/k^2)=-1/k^4\), and the denominator remains \(2/k^4\). Hence

g(1/k,-1/k^2) = \frac{-1/k^4}{2/k^4} = -\frac12.

These function-value sequences converge to different numbers. Consequently, \(g\) has no limit at the origin. The paired sequences follow the curves \(y=x^2\) and \(y=-x^2\), sampled at \(x=1/k\).

Using Sequences to Verify a Limit

The criterion also proves that a limit exists. The key is to start with an arbitrary sequence approaching the point, then establish that its function values must approach the same number. A bound in terms of the distance to the point can make this especially direct.

Worked Example: Every Approaching Sequence Gives the Same Limit

Define \(h(x,y)=\frac{x^2y}{x^2+y^2}\) for \((x,y)\ne(0,0)\). We will show that its limit at the origin is zero using the sequential criterion.

Take any sequence \((x_k,y_k)\ne(0,0)\) that approaches \((0,0)\). Since \(x_k^2\le x_k^2+y_k^2\), and the denominator is positive,

|h(x_k,y_k)| = \frac{x_k^2|y_k|}{x_k^2+y_k^2} \le |y_k| \le \sqrt{x_k^2+y_k^2}.

The final expression is the distance from \((x_k,y_k)\) to the origin, so it tends to zero. Thus \(0\le |h(x_k,y_k)|\le\sqrt{x_k^2+y_k^2}\to0\), which forces \(h(x_k,y_k)\to0\). The sequence was arbitrary, so the criterion proves \(\lim_{(x,y)\to(0,0)}h(x,y)=0\).

As a direct check on the estimate, if \((x,y)=(1/k,1/k)\), then \(h(1/k,1/k)=\frac{1/k^3}{2/k^2}=1/(2k)\), which tends to zero, as the bound predicts.

A Common Mistake: Treating a Few Paths as a Proof

Sequences provide a powerful test, but checking a few sequences cannot establish a multivariable limit. The theorem requires the same function-value limit for every sequence in the domain that approaches the point. A finite collection of successful tests does not cover all possible approaches.

For example, for \(f(x,y)=\frac{xy}{x^2+y^2}\), the coordinate-axis tests both give zero. Along the \(x\)-axis, where \(y=0\) and \(x\ne0\),

f(x,0)=\frac{x\cdot0}{x^2+0}=0.

Along the \(y\)-axis, where \(x=0\) and \(y\ne0\),

f(0,y)=\frac{0\cdot y}{0+y^2}=0.

Concluding from these two calculations that the limit is zero would be incorrect. The diagonal sequence \((1/k,1/k)\) gives the constant value \(1/2\), so the limit does not exist. The correct lesson is asymmetric: one sequence whose function values fail to approach a proposed \(L\) disproves that limit; to prove a limit using the criterion, the claim must hold for every approaching sequence.

Choosing Sequences Carefully

When investigating a suspected failure, look for approaches that simplify the expression but still reach the same point. Straight lines are natural candidates, but if they give the same result, a curved approach may reveal different behavior. Once a sequence pair has been chosen, verify its role rather than relying on appearance:

Check that each sequence term belongs to the function's domain.
Check that the terms are not equal to the point being approached.
Compute the distance to the point and show that it tends to zero.
Substitute the sequence terms into the function and simplify exactly.
Compare the resulting limits; different limits rule out a multivariable limit.

For a proof that a limit exists, the strategy is different: consider an arbitrary sequence approaching the point and show that the corresponding function values converge to the proposed limit. A uniform estimate that applies to all inputs near the point often provides an efficient way to do this.

Key takeaway: A multivariable limit exists with value \(L\) exactly when every sequence of domain points approaching the point produces function values converging to \(L\). Two incompatible sequences are decisive evidence of nonexistence; a few agreeing sequences are not proof of existence.

Check Your Understanding

Use the sequential criterion and the definitions carefully when answering.

What conditions must a sequence satisfy to be used in the criterion for a limit at \(\mathbf{a}\)?
Why does the epsilon-delta definition imply that function values along any approaching sequence converge to \(L\)?
In the function \(f(x,y)=\frac{xy}{x^2+y^2}\), what do the sequences \((1/k,1/k)\) and \((1/k,-1/k)\) show?
Why do matching values along the two coordinate axes not prove that the limit of that function exists?
For \(h(x,y)=\frac{x^2y}{x^2+y^2}\), how does the estimate \(|h(x,y)|\le\sqrt{x^2+y^2}\) establish its limit at the origin?