Continuity Diagnostics for Formula-Defined Functions

Introduction

For a function of several variables, continuity can fail for more than one reason. A formula may not be defined at a point, different approaches to a point may produce incompatible limiting values, or the limit may exist but disagree with the value the function has been assigned.

A reliable diagnosis therefore begins before any limit calculation: identify the function’s domain. Then examine points in that domain, paying particular attention to its boundary and to points where the formula changes or has a potentially problematic denominator.

Diagnostic principle: At a point in the domain, continuity requires the function’s values at nearby domain points to approach the value assigned at the point. When the point is an accumulation point of the domain, this is equivalent to requiring that the joint limit exist and equal the assigned value.

Continuity Relative to a Domain

Let \(f:D\to\mathbb{R}\), where \(D\subseteq\mathbb{R}^2\), and let \(\mathbf{p}=(a,b)\in D\). The function is continuous at \(\mathbf{p}\) if for every \(\varepsilon>0\), there is a \(\delta>0\) such that, whenever \(\mathbf{x}=(x,y)\in D\) and \(\|\mathbf{x}-\mathbf{p}\|<\delta\), then \(|f(\mathbf{x})-f(\mathbf{p})|<\varepsilon\). The condition concerns only points in the domain.

A point \(\mathbf{p}\in D\) is an accumulation point of \(D\) if every open ball centered at \(\mathbf{p}\) contains a point of \(D\) different from \(\mathbf{p}\). At an accumulation point, the limit \(\lim_{\mathbf{x}\to\mathbf{p},\,\mathbf{x}\in D} f(\mathbf{x})\) is defined by considering domain points other than \(\mathbf{p}\).

Definition: If \(\mathbf{p}\in D\) is an accumulation point of \(D\), then \(f\) is continuous at \(\mathbf{p}\) if and only if the limit of \(f(\mathbf{x})\) as \(\mathbf{x}\) approaches \(\mathbf{p}\) through \(D\) exists and equals \(f(\mathbf{p})\). If \(\mathbf{p}\) is isolated in \(D\), then \(f\) is continuous at \(\mathbf{p}\) under the epsilon-delta definition: a sufficiently small ball contains no other domain points, so the condition is vacuously satisfied.

The accumulation-point qualification matters. For example, if \(D=\{(0,0)\}\), the function is continuous at its only point by the epsilon-delta definition. There are no other domain points approaching \((0,0)\), however, so the usual punctured limit through \(D\) is not defined there. It is not correct to use the limit equivalence without checking this condition.

Why the Limit Criterion Works

At an accumulation point, the limit criterion is exactly the continuity definition with the point itself removed from consideration. The proof makes clear where the requirement on the domain enters.

Continuity implies the limit condition.
Suppose \(f\) is continuous at an accumulation point \(\mathbf{p}\). Given \(\varepsilon>0\), continuity supplies \(\delta>0\) such that every \(\mathbf{x}\in D\) with \(\|\mathbf{x}-\mathbf{p}\|<\delta\) satisfies \(|f(\mathbf{x})-f(\mathbf{p})|<\varepsilon\). The same inequality holds for the punctured domain points, so their function values tend to \(f(\mathbf{p})\).

The limit condition implies continuity.
Suppose the limit through \(D\) exists and equals \(f(\mathbf{p})\). For each \(\varepsilon>0\), the limit definition gives a \(\delta>0\) for which \(0<\|\mathbf{x}-\mathbf{p}\|<\delta\), \(\mathbf{x}\in D\), implies \(|f(\mathbf{x})-f(\mathbf{p})|<\varepsilon\). At \(\mathbf{x}=\mathbf{p}\), the difference is \(|f(\mathbf{p})-f(\mathbf{p})|=0<\varepsilon\). Thus the continuity condition holds for every domain point in the ball.

Start with the Domain

For a formula-defined function, determine where each operation makes sense before testing continuity. A quotient requires a nonzero denominator; an even root of a real-valued function requires a nonnegative radicand; and a logarithm requires a positive argument. The domain may exclude points where a simplified expression would have a value.

Continuity is a property of a function at points in its domain. A point excluded from the domain is not, by itself, a discontinuity of that function. It may be useful to ask whether the function has a continuous extension to the missing point, but that is a separate question.

Worked Example: A Missing Diagonal in a Quotient

Consider \(f(x,y)=\dfrac{x^2-y^2}{x-y}\), with domain \(D=\{(x,y)\in\mathbb{R}^2:x\ne y\}\). First, the denominator requires \(x-y\ne0\), so the entire diagonal \(y=x\) is excluded.

For every point in \(D\), factor the numerator and cancel the nonzero factor \(x-y\):

f(x,y)=\frac{(x-y)(x+y)}{x-y}=x+y,\qquad (x,y)\in D.

The polynomial \(x+y\) is continuous, so its restriction to \(D\) is continuous at every point of \(D\). In particular, there is no continuity failure at any point where the original function is defined.

Now consider a point \((c,c)\) on the excluded diagonal. Values from the domain satisfy \(f(x,y)=x+y\), and therefore

\lim_{\substack{(x,y)\to(c,c)\\(x,y)\in D}}f(x,y) = \lim_{\substack{(x,y)\to(c,c)\\(x,y)\in D}}(x+y) = 2c.

The limit exists, but \(f(c,c)\) is not assigned: \((c,c)\notin D\). Thus this point is a hole in the domain, not a point at which the original function is discontinuous. Defining the function there to have value \(2c\) would give the continuous extension \(x+y\), including on the diagonal.

Check: At any domain point, the original quotient equals \(x+y\), so the simplification preserves all its values. At the missing point \((c,c)\), the proposed extension has value \(c+c=2c\), exactly the computed limit.

Joint Limits Must Account for Every Approach

When a point belongs to the domain, checking only a few convenient directions does not establish that a joint limit exists. A path that produces a different limiting value is enough to prove that the limit does not exist. If several paths give the same value, further reasoning is still needed to establish the full joint limit.

Worked Example: Different Paths Give Different Limits

Define \(f:\mathbb{R}^2\to\mathbb{R}\) by

f(x,y)= \begin{cases} \dfrac{xy}{x^2+y^2}, &(x,y)\ne(0,0),\\[4pt] 0, &(x,y)=(0,0). \end{cases}

The assigned value makes the function defined at the origin. Away from the origin, the denominator \(x^2+y^2\) is positive, so the quotient is defined and continuous there. To test continuity at the origin, compare two paths through points other than the origin.

Along \(y=0\), for \(x\ne0\),

f(x,0)=\frac{x\cdot0}{x^2+0^2}=0, \qquad\text{so}\qquad \lim_{x\to0}f(x,0)=0.

Along \(y=x\), for \(x\ne0\),

f(x,x)=\frac{x^2}{x^2+x^2}=\frac12, \qquad\text{so}\qquad \lim_{x\to0}f(x,x)=\frac12.

Both paths approach \((0,0)\), but the resulting limits differ. Consequently, the joint limit at the origin does not exist. Since the function is defined at the origin, it is discontinuous there. This is not a removable discontinuity: no single value assigned at the origin could make two different path limits agree.

Check: On \(y=0\), the numerator is zero and the denominator is \(x^2\), giving \(0\). On \(y=x\), the numerator is \(x^2\) and the denominator is \(2x^2\), giving \(1/2\). The disagreement is exact for every nonzero \(x\) on these paths.

Check Boundary Points and Assigned Values

A boundary point of the domain deserves particular attention because nearby points may approach it only from the part of the plane included in the domain. The relevant limit is always taken through the domain. For a piecewise formula, examine the formulas on the different sides of the boundary, then compare their limiting values with one another and with the value assigned on the boundary.

If the joint limit at a domain point exists but does not equal the assigned value, continuity fails. When the limit exists, changing the value at that point to the limit produces a continuous extension at that point. If the joint limit does not exist, changing only the value at the point cannot repair continuity.

Worked Example: A Piecewise Boundary with One Incorrect Value

Define \(f:\mathbb{R}^2\to\mathbb{R}\) by

f(x,y)= \begin{cases} x+y, & x<0,\\ x^2+y, & x\ge0,\ (x,y)\ne(0,0),\\ 1, & (x,y)=(0,0). \end{cases}

The formula changes across the boundary \(x=0\). At a boundary point \((0,b)\) with \(b\ne0\), the assigned value is \(f(0,b)=b\). To check the joint limit from both sides, write the difference from \(b\) explicitly. For \(x<0\),

|f(x,y)-b|=|x+y-b|\le |x|+|y-b|.

For \(x\ge0\) near \((0,b)\), the point is not the origin and

|f(x,y)-b|=|x^2+y-b|\le x^2+|y-b|.

As \((x,y)\to(0,b)\), both bounds tend to zero. These estimates cover points on both sides and on the boundary, so the joint limit through the domain is \(b\). It equals \(f(0,b)\); hence \(f\) is continuous at every \((0,b)\) with \(b\ne0\).

At the origin, the assigned value is instead \(f(0,0)=1\). For \(x<0\),

|f(x,y)|=|x+y|\le |x|+|y|.

For \(x\ge0\) at points other than the origin,

|f(x,y)|=|x^2+y|\le x^2+|y|.

Each right-hand side tends to zero as \((x,y)\to(0,0)\). Thus the joint limit is zero, whereas the assigned value is one. The function is discontinuous at the origin. Replacing its value there by zero makes the formulas agree continuously across the boundary.

Check: At a boundary point with \(b\ne0\), both side formulas tend to \(b\), the assigned boundary value. At the origin, both side formulas tend to \(0\), which differs from the assigned value \(1\). The failure is therefore caused by disagreement with the assigned value, not by incompatible side limits.

A Practical Diagnostic Sequence

For a formula-defined function, use the following order. It prevents a common error: trying to decide continuity before determining what function and domain are actually specified.

Find the domain.
Record restrictions from denominators, roots, logarithms, and any piecewise conditions. Identify points that are excluded and boundary points where the formula changes.

Check ordinary domain points.
Use continuity laws for elementary functions and their sums, products, quotients with nonzero denominators, and compositions where defined. This often settles all points away from the exceptional set.

Test exceptional points jointly.
At an accumulation point in the domain, test whether the joint limit exists. A pair of paths with different limits disproves existence; a proof of existence must control all approaches.

Compare with the assigned value.
If the limit exists, compare it with the function’s actual value at the point. Equality gives continuity; inequality gives a removable discontinuity when changing that one value supplies the limit.

Common Mistake: Treating Every Missing Point as a Discontinuity

A frequent mistake is to say that a function is discontinuous at every point where its displayed formula is undefined. For example, the quotient in the first worked example is undefined on \(y=x\), but those points are not in its domain. Continuity at a point is only asked for points where the function is defined. One may discuss a possible extension to a missing point, but that does not change the original function’s domain.

A related mistake is to use the limit characterization at an isolated domain point without qualification. The epsilon-delta definition quantifies over domain points near the point; if there are no other such points in a sufficiently small neighborhood, its condition is automatically satisfied. The punctured limit, by contrast, requires an accumulation point. Keeping these statements distinct avoids a false equivalence.

Why These Diagnostics Matter

Continuity checks are not merely a final test after simplifying a formula. They reveal how the function is actually defined and distinguish several different situations: a formula that is regular on its domain, a missing point that may admit an extension, incompatible path behavior that prevents a limit, and a limit that disagrees with an assigned value. These distinctions guide the correct conclusion and determine whether changing a point value can help.

Key takeaway: First identify the domain; then take limits through that domain; finally compare any existing limit with the value assigned at the point. At an isolated domain point, use the epsilon-delta definition rather than an inapplicable punctured-limit equivalence.

Check Your Understanding

For each question, state what the domain or limit information allows you to conclude.

For \(f(x,y)=\dfrac{x+y}{x-y}\), what restriction does the denominator impose on the domain?
Why is a point outside the domain not a point of discontinuity of the original function?
Along two paths to a point, a function approaches \(2\) and \(5\). What can you conclude about its joint limit?
If the joint limit at a domain point exists and equals the assigned value, what does the limit criterion imply, assuming the point is an accumulation point of the domain?
Why is the punctured-limit version of the continuity criterion not suitable at an isolated point of the domain?