Epsilon-Delta Proofs for Multivariable Limits

From Distance to a Limit Proof

A multivariable limit asks what values a function approaches as its input point approaches a specified point. To prove such a claim directly, we must show that making the input sufficiently close to the target point forces the function value to be close to the proposed limit.

The central task is to connect two kinds of distance: the Euclidean distance between input points and the difference between function values. In practice, this means finding an inequality that bounds the function-value error in terms of the distance to the target point. Once that bound is established, it guides an explicit choice of $\delta$.

The Definition and Euclidean Distance

For $\mathbf{x}=(x_1,\ldots,x_n)$ and $\mathbf{a}=(a_1,\ldots,a_n)$ in $\mathbb{R}^n$, their Euclidean distance is the norm of their difference.

\|\mathbf{x}-\mathbf{a}\| = \sqrt{(x_1-a_1)^2+\cdots+(x_n-a_n)^2}.

Suppose $f:D\to\mathbb{R}$, where $D$ is a subset of $\mathbb{R}^n$, and $\mathbf{a}$ is a point whose neighborhoods contain points of $D$ other than possibly $\mathbf{a}$. The limit of $f(\mathbf{x})$ as $\mathbf{x}$ approaches $\mathbf{a}$ is $L$ if every positive tolerance $\varepsilon$ for the output can be guaranteed by choosing a sufficiently small positive radius $\delta$ for the input.

Definition: The statement $\lim_{\mathbf{x}\to\mathbf{a}}f(\mathbf{x})=L$ means that for every $\varepsilon>0$, there exists $\delta>0$ such that, for every $\mathbf{x}\in D$, $$ 0<\|\mathbf{x}-\mathbf{a}\|<\delta \quad\Longrightarrow\quad |f(\mathbf{x})-L|<\varepsilon. $$

The condition $0<\|\mathbf{x}-\mathbf{a}\|$ excludes the point $\mathbf{a}$ itself. A limit concerns nearby inputs, so the function need not even be defined at $\mathbf{a}$. A proof must work for every point in the punctured neighborhood that belongs to the domain, not merely along one selected direction.

How Coordinate Differences Are Controlled

The Euclidean norm contains all the squared coordinate differences. Consequently, no single coordinate difference can exceed the norm. For example, in two variables, if $r=\|\mathbf{x}-\mathbf{a}\|$, then $|x-a_1|\le r$ and $|y-a_2|\le r$. This simple observation often turns an estimate involving separate coordinates into one involving the distance required by the definition.

Start with the output error.
Write $|f(\mathbf{x})-L|$ and transform it until it can be bounded using coordinate differences.

Replace coordinate differences by distance.
Use $|x_i-a_i|\le\|\mathbf{x}-\mathbf{a}\|$, or combine coordinate terms with the Cauchy–Schwarz inequality.

Choose the radius.
Make the resulting distance bound smaller than $\varepsilon$, also imposing any local restriction needed for the estimate.

Check the implication.
Assume $0<\|\mathbf{x}-\mathbf{a}\|<\delta$ and verify explicitly that the output error is less than $\varepsilon$.

Here is the basic coordinate estimate in two variables. Writing $r=\sqrt{(x-a)^2+(y-b)^2}$, we have $(x-a)^2\le r^2$ and $(y-b)^2\le r^2$. Taking nonnegative square roots gives the two coordinate bounds. More generally, for real numbers $c_1,\ldots,c_n$, Cauchy–Schwarz gives a useful bound for a weighted sum of coordinate differences.

|x_i-a_i|\le\|\mathbf{x}-\mathbf{a}\|, \qquad \left|\sum_{i=1}^{n}c_i(x_i-a_i)\right| \le \sqrt{\sum_{i=1}^{n}c_i^2}\,\|\mathbf{x}-\mathbf{a}\|.

The second inequality follows by applying Cauchy–Schwarz to the vectors $(c_1,\ldots,c_n)$ and $(x_1-a_1,\ldots,x_n-a_n)$. It is especially effective when the output error is a linear expression in the coordinate differences.

A General Estimate for Choosing Delta

Suppose an estimate has already shown that, whenever $\mathbf{x}$ is sufficiently close to $\mathbf{a}$, the output error is at most a constant times a power of the input distance. This gives a direct recipe for selecting $\delta$. The proof still needs to respect any neighborhood restriction used to establish the estimate.

Useful estimate: If $C>0$, $p>0$, and $$ |f(\mathbf{x})-L|\le C\|\mathbf{x}-\mathbf{a}\|^p $$ holds whenever $0<\|\mathbf{x}-\mathbf{a}\|<\rho$, then for any $\varepsilon>0$ it is enough to choose $$ \delta=\min\left\{\rho,\left(\frac{\varepsilon}{C}\right)^{1/p}\right\}. $$

Indeed, if $0<\|\mathbf{x}-\mathbf{a}\|<\delta$, then the estimate applies because $\delta\le\rho$. Also, $\|\mathbf{x}-\mathbf{a}\|<(\varepsilon/C)^{1/p}$, so $C\|\mathbf{x}-\mathbf{a}\|^p<\varepsilon$. Thus the defining implication holds. The minimum is important: the estimate may have been proved only in a restricted neighborhood, while the other bound ensures the error meets the requested tolerance.

Worked Examples

Worked Example: A Linear Function of Two Variables

Prove that $\lim_{(x,y)\to(1,-1)}(3x-2y)=5$.

Let $\mathbf{x}=(x,y)$ and $\mathbf{a}=(1,-1)$. The proposed output error is

$$ |(3x-2y)-5| =|3(x-1)-2(y+1)|. $$

By Cauchy–Schwarz, this is at most the length of the coefficient vector $(3,-2)$ times the Euclidean distance from $(x,y)$ to $(1,-1)$:

|3(x-1)-2(y+1)| \le \sqrt{3^2+(-2)^2}\sqrt{(x-1)^2+(y+1)^2} =\sqrt{13}\,\|\mathbf{x}-\mathbf{a}\|.

Given any $\varepsilon>0$, choose $\delta=\varepsilon/\sqrt{13}$. If $0<\|\mathbf{x}-\mathbf{a}\|<\delta$, then

|(3x-2y)-5| \le\sqrt{13}\,\|\mathbf{x}-\mathbf{a}\| <\sqrt{13}\,\delta =\varepsilon.

This verifies the required implication for every $\varepsilon>0$, so the limit is $5$.

Worked Example: A Polynomial at a Nonzero Point

Prove that $\lim_{(x,y)\to(1,2)}(x^2+y^2)=5$.

Factor each difference of squares and set $r=\|\mathbf{x}-(1,2)\|$. If $r<1$, then $|x-1|<1$ and $|y-2|<1$. Therefore $|x+1|\le |x-1|+2<3$, and $|y+2|\le |y-2|+4<5$. It follows that

\begin{aligned} |x^2+y^2-5| &=|(x-1)(x+1)+(y-2)(y+2)|\\ &\le |x-1||x+1|+|y-2||y+2|\\ &<3|x-1|+5|y-2|\\ &\le \sqrt{34}\sqrt{(x-1)^2+(y-2)^2} =\sqrt{34}\,r. \end{aligned}

The last inequality is Cauchy–Schwarz applied to $(3,5)$ and $(|x-1|,|y-2|)$. For a given $\varepsilon>0$, choose $\delta=\min\{1,\varepsilon/\sqrt{34}\}$. If $0<r<\delta$, then $r<1$, so the estimate applies, and $\sqrt{34}\,r<\varepsilon$. Hence $|x^2+y^2-5|<\varepsilon$, proving the limit.

As a check, the expression for the error is exactly $x^2+y^2-5$, which becomes zero at the target point $(1,2)$; the estimate above controls its size at nearby points rather than relying on that substitution alone.

Worked Example: A Product at the Origin

Prove that $\lim_{(x,y)\to(0,0)}xy=0$.

Let $r=\sqrt{x^2+y^2}$. Since $|x|\le r$ and $|y|\le r$, multiplying these nonnegative inequalities gives $|xy|\le r^2$. Given $\varepsilon>0$, choose $\delta=\sqrt{\varepsilon}$. Whenever $0<r<\delta$,

|xy|\le r^2<\delta^2=\varepsilon.

This is precisely the required output bound, so the limit is zero. In this example, the error is controlled by the square of the input distance, which is why the square root of $\varepsilon$ is a suitable radius.

Worked Example: A Square-Root Expression

Prove that $\lim_{(x,y)\to(0,0)}\sqrt{1+x^2+y^2}=1$.

Let $r=\sqrt{x^2+y^2}$. The expression is defined for every $(x,y)\in\mathbb{R}^2$. Rationalizing the output error gives

\left|\sqrt{1+x^2+y^2}-1\right| = \frac{x^2+y^2}{\sqrt{1+x^2+y^2}+1} = \frac{r^2}{\sqrt{1+r^2}+1} \le \frac{r^2}{2},

because $\sqrt{1+r^2}\ge 1$. Given $\varepsilon>0$, choose $\delta=\sqrt{2\varepsilon}$. If $0<r<\delta$, then

\left|\sqrt{1+x^2+y^2}-1\right| \le\frac{r^2}{2} <\frac{\delta^2}{2} =\varepsilon.

The bound verifies the definition, and the proposed limit is $1$.

A Common Pitfall: Choosing Delta Without a Bound

A choice of $\delta$ must be justified by an inequality. It is not enough that $\delta$ seems small or that the function appears close to its proposed limit. For the polynomial example, choosing $\delta=\varepsilon$ without further adjustment does not establish the required bound.

For instance, take $\varepsilon=1$ and the point $(x,y)=(1.9,2)$. Its distance from $(1,2)$ is $0.9<1=\delta$, but the output error is

$$ |1.9^2+2^2-5| =|3.61+4-5| =2.61>1. $$

Thus that proposed choice fails. The correction is to use the estimate established in the proof and choose $\delta=\min\{1,\varepsilon/\sqrt{34}\}$. Then the distance condition forces the output error to be less than $\varepsilon$. The lesson is general: first prove a valid bound, then choose $\delta$ from that bound, and finally check the implication.

Why Explicit Epsilon-Delta Proofs Matter

An explicit proof makes the logical structure of a limit claim visible. The arbitrary $\varepsilon>0$ represents any accuracy demanded of the output; the chosen $\delta$ shows how close the input must be to meet that demand. A norm estimate is especially useful because it handles all directions of approach at once: every coordinate displacement is constrained by the same Euclidean distance.

Different problems may require different estimates. A linear error can often be bounded directly by a constant times the distance. A quadratic error may be bounded by a constant times the square of the distance. A local estimate may require first restricting the distance to be less than some fixed radius. In each case, the proof is complete only when the chosen $\delta$ works for every domain point satisfying the punctured-distance condition.

Check Your Understanding

For each question, identify the distance estimate or logical condition that makes the proof work.

State the epsilon-delta definition of $\lim_{\mathbf{x}\to\mathbf{a}}f(\mathbf{x})=L$, including the role of the punctured neighborhood.
Why does $|x_i-a_i|\le\|\mathbf{x}-\mathbf{a}\|$ hold for each coordinate?
If $|f(\mathbf{x})-L|\le 4\|\mathbf{x}-\mathbf{a}\|^2$, with no additional local restriction, what choice of $\delta$ is sufficient for a given $\varepsilon>0$?
In the polynomial example at $(1,2)$, why is it useful to include $1$ in the minimum defining $\delta$?
Explain why a choice of $\delta$ must be checked against the output error rather than judged only by whether it is positive and small.