When a Missing Value Can Be Repaired
A formula can fail to define a function at a point even when its nearby values follow a clear pattern. For example, a quotient may have both numerator and denominator equal to zero at one point, although cancellation shows that the quotient agrees with a simpler expression everywhere else in its domain.
In one or several variables, the key question is whether the function values approach a single finite number as the input approaches the excluded point. If they do, assigning that number at the point repairs the discontinuity and makes the function continuous there. If the joint limit does not exist, no choice of value at the point can make the function continuous.
Definition of a Continuous Extension
Let \(f\) be defined on a set \(D\) containing points arbitrarily close to a point \(\mathbf{a}\), but not containing \(\mathbf{a}\) itself. For two variables, write \(\mathbf{a}=(a,b)\) and use the distance \(\sqrt{(x-a)^2+(y-b)^2}\) to measure how close \((x,y)\) is to \((a,b)\).
The word joint matters: the input approaches the point through the domain in any allowed direction, not merely along one selected curve. A limit along a particular path can help test whether the joint limit exists, but a single matching path does not establish it.
Continuity at \(\mathbf{a}\) means that \(F(\mathbf{a})\) is defined and \(\lim_{\mathbf{x}\to\mathbf{a}}F(\mathbf{x})=F(\mathbf{a})\). Thus, when the original function has a finite joint limit \(L\), choosing \(F(\mathbf{a})=L\) is not just a convenient choice: it is the only value that can make the extension continuous.
Worked Example: Cancel a Factor and Extend Across a Line
Worked Example: Cancel a Factor and Extend Across a Line
Consider \(f(x,y)=\frac{x^2-y^2}{x-y}\), defined only when \(x\ne y\). Determine whether it can be extended continuously to the excluded line, and find the extension value at an arbitrary point \((a,a)\) on that line.
For every point in the original domain, factor the numerator:
Since \(x-y\ne 0\) on the domain, division by \(x-y\) is valid there, so
As \((x,y)\to(a,a)\) through the original domain, the expression \(x+y\) tends to \(a+a=2a\). Therefore, the joint limit is \(2a\), and the required value at \((a,a)\) is \(F(a,a)=2a\).
In fact, the formula for the extension is \(F(x,y)=x+y\) on the entire plane: off the line it agrees with \(f\), and on the line \(x=y\) its value is \(2x\). To check continuity at \((a,a)\), let \(h=x-a\) and \(k=y-a\). Then
Given any \(\varepsilon>0\), choosing \(\delta=\varepsilon/\sqrt{2}\) makes the final quantity less than \(\varepsilon\) whenever \(0<\sqrt{h^2+k^2}<\delta\). This verifies continuity. The extension is exactly the polynomial \(x+y\), so it is continuous everywhere.
Why the Limit Determines the Extension
The extension rule follows directly from the definition of continuity. Suppose \(f\) is defined near, but not at, \(\mathbf{a}\), and suppose its finite joint limit there is \(L\). Define \(F\) to equal \(f\) at all original domain points and set \(F(\mathbf{a})=L\). The limit of \(F\) as \(\mathbf{x}\to\mathbf{a}\) is still \(L\), because away from \(\mathbf{a}\), \(F\) has exactly the same values as \(f\). Hence
which is the definition of continuity at \(\mathbf{a}\). Conversely, if any extension \(G\) is continuous at \(\mathbf{a}\), then
The first equality holds because \(f\) and \(G\) agree at every nearby point where the original function is defined. Therefore, the joint limit must exist and the assigned value must be that limit. This proves both that the limit condition is sufficient and that the extension value is unique.
Worked Example: Bound the Formula by Distance
Worked Example: Bound the Formula by Distance
Let \(f(x,y)=\frac{x^4+y^4}{x^2+y^2}\) for \((x,y)\ne(0,0)\). Find the limit at the origin and determine the continuous extension value there.
For all real \(x,y\), both \(x^4\) and \(y^4\) are nonnegative, and
For \((x,y)\ne(0,0)\), the denominator \(x^2+y^2\) is positive. Dividing the inequality by it gives
As \((x,y)\to(0,0)\), the upper bound \(x^2+y^2\) tends to zero. The squeeze theorem therefore gives
Thus the unique continuous extension assigns \(F(0,0)=0\). For a direct continuity check, \(F\) agrees with \(f\) away from the origin and satisfies \(0\le F(x,y)\le x^2+y^2\); at the origin it equals zero. Given \(\varepsilon>0\), if \(\sqrt{x^2+y^2}<\sqrt{\varepsilon}\), then \(|F(x,y)-F(0,0)|\le x^2+y^2<\varepsilon\). This confirms continuity at the origin.
When No Value Can Repair the Discontinuity
A missing value alone does not guarantee a removable discontinuity. If nearby function values approach different numbers along different paths, there is no joint limit to assign. Earlier material on path tests gives a practical way to detect this failure: two paths approaching the same point with different limiting values rule out a joint limit.
Worked Example: Show That No Extension Exists
Consider \(f(x,y)=\frac{xy}{x^2+y^2}\), defined for \((x,y)\ne(0,0)\). Decide whether some value assigned at the origin can make the function continuous.
Approach the origin along the line \(y=x\), with \(x\ne0\). Substitution gives
The limit along this path is \(1/2\). Now approach along \(y=-x\), again with \(x\ne0\):
This path has limit \(-1/2\). Both paths approach \((0,0)\), but their function values approach different numbers. Therefore, the joint limit does not exist. If an extension assigned a value \(c\) at the origin and were continuous there, every path limit would have to equal \(c\). Since it would have to equal both \(1/2\) and \(-1/2\), no such value exists.
A Common Mistake: Treating Zero Over Zero as a Value
When the numerator and denominator both vanish at the excluded point, it is incorrect to treat the resulting \(0/0\) as the function value, or to assume that assigning zero will repair the discontinuity. The expression \(0/0\) is undefined; it is not a number and does not determine a limit.
For the last example, a mistaken calculation might say: “At the origin the numerator is \(0\) and the denominator is \(0\), so set \(f(0,0)=0\).” The correction is to examine nearby values. Along \(y=x\), they are \(1/2\); along \(y=-x\), they are \(-1/2\). Thus, setting the value to zero does not produce continuity, and neither does any other assignment.
Cancellation also requires care. In the first example, cancelling \(x-y\) is valid only where \(x-y\ne0\), which is precisely the original domain. The simplified formula \(x+y\) describes the values there and suggests an extension, but it does not retroactively make the original quotient defined on the excluded line. The extended function is a new function whose definition explicitly supplies those missing values.
A Reliable Decision Process
For a formula with an excluded point, separate the question of the limit from the question of the assigned value. A useful sequence is:
Check where the formula is actually defined; do not assume that a simplified expression has the same domain.
Use algebraic simplification where valid, or a bound that forces the function values toward one finite number. Different path limits show that no joint limit exists.
Set the extension's value at the excluded point equal to the finite joint limit. No other value can give continuity.
Confirm that the limit of the extended function equals its assigned value at the point.
Why Continuous Extensions Matter
A removable discontinuity records a defect in how a formula was originally written or restricted, rather than an unavoidable failure of nearby values to settle down. Replacing a quotient by a simpler expression on its original domain can reveal the missing value. Defining that value carefully gives a continuous function without changing any of the original function's values.
This distinction is useful whenever formulas are defined by cases, contain denominators that vanish at particular inputs, or arise from expressions that simplify away from an exceptional set. The extension must be justified by the relevant joint limit; visual simplicity or a convenient assignment is not enough. In particular, continuity is a local condition at each point being filled in: if extending across an entire excluded set, the limit and assigned value must agree at every point of that set.
Check Your Understanding
Use the definition of a continuous extension and the joint limit at the excluded point in each question.
- For \(f(x,y)=\frac{x^2-y^2}{x-y}\) when \(x\ne y\), what value must an extension assign at \((3,3)\)?
- Why does the existence of a finite joint limit guarantee a continuous extension when the value at the point is set equal to that limit?
- For \(f(x,y)=\frac{x^4+y^4}{x^2+y^2}\) away from the origin, state an inequality that proves its limit at the origin is zero.
- For \(f(x,y)=\frac{xy}{x^2+y^2}\) away from the origin, what do the paths \(y=x\) and \(y=-x\) show about the possibility of extension?
- Why is it not valid to use the value of a simplified formula at an excluded point as the original function's value without defining an extension?