Why Describe Limits with Neighborhoods?
A multivariable function can approach a point from many directions. In two dimensions, points can come from anywhere in the plane; in three dimensions, they can come from anywhere in space. A useful definition must account for all nearby domain points, not just a chosen direction.
Neighborhoods provide a precise way to describe which inputs count as nearby. A punctured neighborhood includes nearby points but excludes the center itself. That exclusion matters: a limit describes the function’s behavior around a point, while the function’s value at the point is a separate piece of information.
Distance and Neighborhoods
For points in two or three dimensions, distance is measured with the Euclidean norm. If \(\mathbf{x}=(x,y)\) and \(\mathbf{a}=(a_1,a_2)\) are points in the plane, their distance is \(\|\mathbf{x}-\mathbf{a}\|\). In space, the same notation applies to points with three coordinates.
An open ball of radius \(r>0\) centered at \(\mathbf{a}\) consists of all points whose distance from \(\mathbf{a}\) is less than \(r\). A punctured ball is this ball with its center removed.
In the plane, \(B_r(\mathbf{a})\) is a disk; in space, it is a solid ball. In either case, the punctured version removes only the center, not any of the surrounding points.
The point \(\mathbf{a}\) may or may not belong to \(D\). If it does belong to \(D\), the limit still examines the other domain points near \(\mathbf{a}\), not the value at the center. Requiring an accumulation point prevents the definition from making a claim based on a punctured neighborhood containing no domain points.
The Neighborhood Definition of a Limit
For a scalar-valued function, an output neighborhood of a proposed limit \(L\) is an open interval centered at \(L\). Its radius describes how close function values must be to \(L\). The input neighborhood can then be chosen so that every domain point in its punctured part has a function value inside that output interval.
Here, \(s\) is the radius of the output neighborhood around \(L\), and \(r\) is the radius of an input neighborhood around \(\mathbf{a}\). The key requirement is that every domain point in the chosen punctured input neighborhood has a function value in the required output neighborhood. The choice of \(r\) may depend on \(s\), but it must work for all such inputs at once.
The norm makes the same definition work in either dimension. In two dimensions, the input distance is \(\sqrt{(x-a_1)^2+(y-a_2)^2}\). In three dimensions, it is \(\sqrt{(x-a_1)^2+(y-a_2)^2+(z-a_3)^2}\). The domain condition matters: only inputs belonging to \(D\) are tested.
Worked Examples with Neighborhoods
Worked Example: A Limit in the Plane
Let \(f(x,y)=x+y\), with domain \(D=\mathbb{R}^2\). Show from the neighborhood definition that the limit at \((0,0)\) is \(0\).
Take any output radius \(s>0\). Choose the input radius \(r=s/2\). If \((x,y)\in D\) and \(0<\sqrt{x^2+y^2}<r\), then \(|x|<r\) and \(|y|<r\), because each coordinate’s absolute value is at most the distance from the origin. Therefore,
Thus every domain point in that punctured input ball has a function value in the output interval of radius \(s\) around \(0\). Since this works for every \(s>0\), the limit is \(0\). Checking the chosen radius gives \(2r=2(s/2)=s\), as required.
Worked Example: A Limit in Space
Let \(f(x,y,z)=z\), with domain \(D=\mathbb{R}^3\), and consider the point \(\mathbf{a}=(1,-1,2)\). Show that the limit there is \(2\).
Take any output radius \(s>0\), and choose the input radius \(r=s\). For any \((x,y,z)\) satisfying \(0<\|(x,y,z)-(1,-1,2)\|<r\), the absolute change in the third coordinate cannot exceed the full distance. Hence,
This places every such function value within the output neighborhood of radius \(s\) around \(2\). The chosen input radius works for each output radius, so the limit is \(2\). The check is the displayed inequality: its right-hand side is strictly less than \(s\), exactly the required output condition.
Worked Example: The Limit Can Differ from the Value
Define \(f:\mathbb{R}^2\to\mathbb{R}\) by \(f(x,y)=x^2+y^2\) when \((x,y)\ne(0,0)\), and set \(f(0,0)=9\). Find the limit at the origin and compare it with the function value there.
For any output radius \(s>0\), choose the input radius \(r=\sqrt{s}\). If \(0<\sqrt{x^2+y^2}<r\), then
Therefore the limit at the origin is \(0\). This conclusion concerns only inputs in the punctured ball, so the assigned center value does not enter the calculation. Directly from the definition of the function, \(f(0,0)=9\). Thus the limiting value is \(0\), while the function value at the point is \(9\). Substituting the chosen \(r\) back into the bound gives \(r^2=(\sqrt{s})^2=s\), confirming the required neighborhood inclusion.
Why the Center Is Excluded
The condition \(0<\|\mathbf{x}-\mathbf{a}\|\) excludes exactly the center. It does not exclude points merely because they are close to the center. This is what allows a limit to describe nearby behavior independently of the value at \(\mathbf{a}\).
In particular, if two functions agree at every domain point except possibly \(\mathbf{a}\), then their limits at \(\mathbf{a}\), when considered, have the same neighborhood behavior. Every punctured input neighborhood omits \(\mathbf{a}\), so changing the value there cannot alter any of the function values tested by the definition. A limit can therefore exist even if the function is not defined at the point, and a defined function value need not equal its limit.
A Limit, If It Exists, Is Unique
The neighborhood definition cannot assign two different limits to the same function at the same point. Here is why. Suppose the limits were \(L_1\) and \(L_2\), with \(L_1\ne L_2\), and let \(d=|L_1-L_2|>0\). The output intervals of radius \(d/3\) centered at \(L_1\) and \(L_2\) are disjoint: if a number lay in both, the distance between the centers would be less than \(2d/3\), contradicting that their distance is \(d\).
By the definition of the first proposed limit, some punctured input ball has all its function values in the interval around \(L_1\). By the definition of the second, some punctured input ball has all its function values in the interval around \(L_2\). Choose the smaller of the two input radii. Since \(\mathbf{a}\) is an accumulation point of \(D\), this smaller punctured ball contains a domain point \(\mathbf{x}\). Its function value would have to lie in both disjoint output intervals, which is impossible. Thus two distinct limits cannot exist.
Common Mistake: Including the Center
A common error is to test a function’s value at the center as if it were one of the nearby values required by the limit definition. For the function in the third worked example, this incorrect calculation would be
The conclusion does not follow: it uses the value at the very point that the punctured neighborhood excludes. The correction is to test inputs with \(0<\sqrt{x^2+y^2}<r\). For those inputs, \(f(x,y)=x^2+y^2\), and the earlier bound shows that their values approach \(0\). The function value \(9\) and the limit \(0\) are different quantities.
How to Read the Definition
Fix any radius \(s>0\) around the proposed limiting value \(L\).
Choose a radius \(r>0\) around \(\mathbf{a}\) that makes the function values close enough.
Every \(\mathbf{x}\in D\) with \(0<\|\mathbf{x}-\mathbf{a}\|<r\) must satisfy \(|f(\mathbf{x})-L|<s\).
The quantifier “every” is essential. It is not enough that some nearby inputs have values near \(L\); all domain inputs in a sufficiently small punctured neighborhood must meet the output requirement. In two and three dimensions, the distance condition includes every direction around the point automatically.
Check Your Understanding
Use the definitions and examples above to answer these questions.
- What points belong to \(B_r^*(\mathbf{a})\), and which point is excluded?
- Why must the center be an accumulation point of the domain when defining a limit there?
- In the limit definition, what roles do the output radius \(s\) and input radius \(r\) play?
- Can a function have a limit at a point where it is not defined? Explain using the punctured-neighborhood definition.
- For the function with \(f(0,0)=9\) and \(f(x,y)=x^2+y^2\) away from the origin, what is the limit at the origin, and why is it not \(f(0,0)\)?