Calc 3: Why is delta the radius of a circle in the epsilon-delta definition of a multivariable limit?

Question

The single variable definition of a limit states (assuming $\epsilon > 0$ and $\delta > 0$) that if $0 < |x-a| < \delta$ then $|f(x) - L| < \epsilon$. The multivariable definition goes if $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$ then $|f(x,y) - L| < \epsilon$. I understand how to use the formula, but I'm just trying to understand from an intuitive standpoint why the multivariable version essentially has a circle ($\sqrt{(x-a)^2 + (y-b)^2}$)?

Answer 1

The notion of limit requires the concept of distance, in some sense - if one must describe that some value approaches another value as some parameter changes, then it is necessary to be precise about what something approaching something does actually mean.

In the real line, the distance between two points is thought of as the absolute value of the subtraction of one point by another. In the plane, there is one extra dimension, so one must come up with some other way to define distance.

One way to do it is using Pythagoras' Theorem: given two points $(x_1,y_1)$ and $(x_2,y_2)$, it is possible to construct a right triangle adding the point $(x_2,y_1)$. The sizes of the legs of the triangle are easy to calculate, using the distance defined on the line: they are $|x_1 - x_2|$ and $|y_1 - y_2|$. Using the theorem, the hypothenuse has length equal to $\sqrt{|x_1 - x_2|^2 + |y_1 - y_2|^2}$.

Using this ideia of distance, one can fix a point and describe the set of points that are some specific distance away from it, or less than that value. For example, the set of points that are less than $\delta$ away from $(a,b)$ is precisely $$\biggl\{ (x,y) \in \mathbb{R}^2 : \sqrt{(x - a)^2 + (y - b)^2} < \delta \biggr\}.$$ To describe some point approaching another point, these are the kind of sets that are used, as in the definition of limit in the plane.

As others have already said, there are other ways of measuring distance in the plane: the distance between $(x,y)$ and $(a,b)$ can be $|x - a| + |y - b|$, and then the set of points that satisfy $|x - a| + |y - b| < \delta$ forms a diamond (a rotated square), or it can be the greater value between $|x - a|$ and $|y - b|$, which is represented by $\max{\{ |x - a|, |y - b| \}}$, and the set of points that satisfy $\max{\{ |x - a|, |y - b| \}} < \delta$ is a square.

These distances are somewhat equivalent because one can fit circles inside squares and squares inside circles: for example, the set $$\{ (x,y) \in \mathbb{R}^2 : |x - a| + |y - b| < \delta \}$$ is inside the above circle and the circle is inside the set $$\{ (x,y) \in \mathbb{R}^2 : |x - a| + |y - b| < \delta \sqrt{2} \}.$$ This means that studying smaller and smaller circles around some fixed point and smaller and smaller squares around some fixed point gets you to the same place - the "approaching" each one defines is the same.

These ideias are all formalized and generalized in the theory of metric spaces, which is a very important area of mathematics.

Answer 2

When we say that $$\lim\limits_{x\to c}f(x)=L$$ we use the definition $$\forall\epsilon\gt 0\text{ there exists }\delta\gt 0\text{ such that }\lvert x-c\rvert\lt\delta\to\lvert f(x)-L\rvert\lt\epsilon$$ to determine if the limit exists and is equal to $L$. What are we really saying?

We are saying that the limit exists and is equal to $L$ if no matter what positive distance $\epsilon$ we specify, if $x$ is within a distance of $\delta$ to $c$, then $f(x)$ will be within a distance of $\epsilon$ to $L$.

Suppose we want the same definition for a function $f(x,y)$. We want a way to say that if an element of the domain $(x,y)$ is within a distance $\delta$ of $(a,b)$ then $f(x,y)$ will be within a distance of $L$.

The distance in two dimensions is given by the Pythagorean theorem so we say $$\lim\limits_{(x,y)\to (ab)}f(x,y)=L$$ if $$\forall\epsilon\gt 0\text{ there exists }\delta\gt 0\text{ such that }\sqrt{(x-a)^2+(y-b)^2}\lt\delta\to\lvert f(x)-L\rvert\lt\epsilon$$