From a Formula to Its Continuity Domain
A multivariable formula can be continuous wherever it is defined, yet it may not be defined at every point in the plane. To find its continuity domain, first determine which inputs make every operation in the formula meaningful as a real number. Then use the continuity of the elementary operations and the continuity laws for sums, products, quotients, and compositions.
The restrictions can interact. A square root may allow its radicand to be zero, while the same square root in a denominator must be strictly positive. A logarithm requires a positive input, and a tangent requires its cosine denominator to be nonzero. The full domain is where all the relevant conditions hold at once.
Continuity at a point of a domain is understood relative to that domain: inputs approaching the point are restricted to points where the function is defined. In particular, a point on the boundary of a domain can still be a point of continuity.
Basic Domain Restrictions
Each type of operation supplies its own condition. When an expression combines several operations, keep all the conditions; do not replace them with just the most visible one.
| Expression or operation | Real-domain condition |
|---|---|
| Polynomial in \(x,y,\ldots\) | All real inputs |
| Quotient \(P/Q\) | \(Q\ne 0\) |
| Even root \(\sqrt[2m]{R}\) | \(R\ge 0\) |
| Even root in a denominator, \(1/\sqrt[2m]{R}\) | \(R>0\) |
| Logarithm \(\ln R\) | \(R>0\) |
| \(\sin R\), \(\cos R\) | No additional restriction |
| \(\tan R\), \(\sec R\) | \(\cos R\ne 0\) |
| \(\cot R\), \(\csc R\) | \(\sin R\ne 0\) |
| \(\arcsin R\), \(\arccos R\) | \(-1\le R\le 1\) |
| \(\arctan R\) | No additional restriction |
Here \(P,Q,R\) may themselves be expressions in several variables, so their own domain restrictions must also be satisfied. For example, in \(\ln(\sqrt{R})\), the logarithm requires \(\sqrt{R}>0\), which is equivalent to \(R>0\), not merely \(R\ge 0\).
Why the Restrictions Combine by Intersection
Suppose a formula contains two subexpressions. The first is defined on a set \(A\), and the second is defined on a set \(B\). The full formula can be evaluated only at inputs where both subexpressions are defined. Its domain is therefore \(A\cap B\), possibly with additional conditions imposed by the operation joining them.
For a quotient, for instance, it is not enough that the numerator and denominator are each defined: the denominator must also be nonzero. For a logarithm of an expression, the inner expression must first be defined, and its value must then be positive.
To justify this, begin with polynomials, which are continuous everywhere. Sums and products of continuous functions are continuous wherever both functions are defined. A quotient of continuous functions is continuous at points where its denominator is nonzero. Composing continuous functions preserves continuity at points where the inner function’s values lie in the outer function’s domain. The square-root, logarithmic, and trigonometric functions are continuous on their respective real domains. Applying these facts as the formula is built shows that it is continuous wherever all its operations are defined.
This reasoning explains why the domain calculation is the central task: once the full natural domain has been correctly identified, continuity follows from the continuity laws. The formula need not be defined on an open set for this conclusion to hold.
Worked Example: A Root Divided by a Logarithm
Worked Example: A Root Divided by a Logarithm
Find the natural domain and continuity domain of
Step 1: Apply the square-root restriction. The radicand must be nonnegative:
Step 2: Apply the logarithm restriction. Its input must be positive:
Step 3: Exclude zeros of the denominator. The logarithm is the denominator, so it cannot equal zero. Since \(\ln u=0\) exactly when \(u=1\), we require
Combining the three conditions gives the domain:
At \((3,0)\), the radicand is \(9-9-0=0\), the logarithm’s input is \(5-3-0=2>0\), and \(\ln 2\ne 0\). Thus \((3,0)\) is in the domain, and
At \((2,2)\), the radicand is \(9-4-4=1\), but the logarithm’s input is \(5-2-2=1\), so the denominator is \(\ln 1=0\). Therefore \((2,2)\) is excluded. At every point that satisfies all three domain conditions, the component operations and quotient are continuous. Hence the continuity domain is exactly the displayed natural domain.
Worked Example: Tangent over a Square Root
Worked Example: Tangent over a Square Root
Find the domain and continuity domain of
The square root is in the denominator. It must be defined and nonzero, so its radicand must be strictly positive:
The tangent is defined when its cosine denominator is nonzero. Since \(\cos t=0\) at \(t=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb Z\), the second condition is
Thus the domain is
At \((1,1)\), the radicand is \(2(1)+1-1=2>0\), and \(x-y=0\), for which \(\cos 0=1\ne 0\). The point is in the domain, and direct substitution gives
The point \((\frac12,0)\) fails the root condition because \(2(\frac12)+0-1=0\); the denominator would be zero. A point with \(x-y=\frac{\pi}{2}\) also fails the tangent condition if it meets the root condition. For example, at \((\frac{\pi}{2},0)\), the radicand is \(\pi-1>0\), but \(\tan(\frac{\pi}{2})\) is undefined. Both restrictions are necessary. By continuity of tangent on its domain and of the square root where its argument is positive, \(G\) is continuous throughout the domain just found.
Worked Example: A Trigonometric Function on a Disk
Worked Example: A Trigonometric Function on a Disk
Determine the domain and continuity domain of
The square root requires
Sine accepts every real input, so it adds no restriction once the square root is defined. The denominator must be nonzero:
Therefore, the domain is the closed disk \(x^2+y^2\le 4\) with the circle \(x^2+y^2=1\) removed. At \((2,0)\), the root’s radicand is \(4-4=0\), and the denominator is \(4-1=3\). This point is included, and
This calculation checks that the boundary point is permitted: the root may equal zero, and the denominator does not. By contrast, \((1,0)\) is excluded because its denominator is \(1-1=0\), even though its root is defined. The function is continuous at every point of its domain, including the included boundary of the disk.
A Common Mistake: Treating Every Root Condition the Same
A frequent error is to use a non-strict inequality for a square root that appears in a denominator. For example, the incorrect calculation for
might be: “The radicand must be nonnegative, so \(x^2+y^2\le 1\).” That condition only ensures that the square root is defined. On the boundary \(x^2+y^2=1\), its value is zero, so the quotient is undefined. The correct condition is
The converse error is also possible: excluding a zero radicand when the root is not in a denominator. In the third example, the boundary \(x^2+y^2=4\) is allowed because the numerator remains defined there and the denominator is nonzero.
A Reliable Procedure
For a formula involving several variables, apply the following steps in order. Keeping each restriction visible makes it easier to detect a missing condition.
List every quotient, even root, logarithm, and trigonometric function whose domain is restricted.
Use a nonzero denominator, a nonnegative radicand for a root, a positive logarithm input, and the appropriate trigonometric exclusions.
Check that each inner expression is defined before applying the outer operation’s condition.
The natural domain is the set of inputs satisfying every condition simultaneously.
Use continuity of the elementary operations and the continuity laws to conclude continuity at every point of that domain.
Finding these domains matters because it identifies where a formula genuinely defines a continuous function, and where an attempted substitution is not even meaningful. It also prevents a boundary point from being discarded automatically: whether it belongs depends on the operations in the formula, not merely on its location at the edge of a region.
Check Your Understanding
For each question, state the restrictions that determine the real domain before deciding whether the function is continuous there.
- What is the domain of \(f(x,y)=\ln(9-x^2-y^2)\)?
- For \(g(x,y)=\frac{\sqrt{x+y}}{x-y}\), which points on \(x+y=0\) can belong to its domain?
- Why must the condition for \(\frac{1}{\sqrt{R(x,y)}}\) be stricter than the condition for \(\sqrt{R(x,y)}\)?
- What inputs must be excluded from \(\tan(x+y)\)?
- Can a formula of this type be continuous at a point on the boundary of its natural domain? Explain briefly.