Continuity at a Point in Several Variables

What Continuity at a Point Requires

A function of several variables can approach a value as its input approaches a point. Continuity at that point requires more than the existence of this joint limit: the function must also be defined at the point, and its value there must agree with the limit.

This distinction is especially useful for functions given by different formulas in different parts of their domain. At a point where the formula changes, evaluate the function using the branch that actually includes the point, then compare that value with the joint limit.

Central test: At a point where the joint limit is defined, a function is continuous precisely when its value at the point exists and equals that limit.

Definition of Continuity

Let \(D\) be a subset of \(\mathbb{R}^2\), let \(f:D\to\mathbb{R}\), and let \((a,b)\in D\). Continuity is defined by comparing function values at nearby domain points with \(f(a,b)\). In the usual case, where \((a,b)\) is an accumulation point of \(D\), this is equivalent to saying that the joint limit equals the function value.

Definition: The function \(f\) is continuous at \((a,b)\) if, for every \(\varepsilon>0\), there exists \(\delta>0\) such that whenever \((x,y)\in D\) and \(\sqrt{(x-a)^2+(y-b)^2}<\delta\), we have \(|f(x,y)-f(a,b)|<\varepsilon\).

The distance \(\sqrt{(x-a)^2+(y-b)^2}\) measures how close \((x,y)\) is to \((a,b)\). The definition says that input points sufficiently close to \((a,b)\) have function values as close as desired to the particular value \(f(a,b)\).

If \((a,b)\) is an accumulation point of \(D\), the definition can be stated in terms of the joint limit:

f \text{ is continuous at } (a,b) \quad\Longleftrightarrow\quad \lim_{(x,y)\to(a,b)} f(x,y)=f(a,b).

The limit concerns values at points approaching \((a,b)\), not the value at the point itself. The right-hand side therefore requires both that the limit exist and that the function have a value at \((a,b)\). If the point is isolated in \(D\), the epsilon-delta definition is automatically satisfied: a sufficiently small neighborhood contains no other domain points. The limit-based test is generally used when the point is an accumulation point.

Why the Limit-and-Value Test Is Equivalent

For an accumulation point \((a,b)\), the epsilon-delta definition of the limit \(\lim_{(x,y)\to(a,b)}f(x,y)=L\) requires that for every \(\varepsilon>0\), some \(\delta>0\) makes \(|f(x,y)-L|<\varepsilon\) whenever \((x,y)\in D\) and \(0<\sqrt{(x-a)^2+(y-b)^2}<\delta\). The definition of continuity uses the same condition with \(L=f(a,b)\). The center itself causes no difficulty in the continuity definition, since \(|f(a,b)-f(a,b)|=0<\varepsilon\). This gives the equivalence directly.

Find the defined value.
Evaluate \(f(a,b)\) using the rule that applies at \((a,b)\).

Find the joint limit.
Determine whether the function values approach a single number as \((x,y)\) approaches \((a,b)\).

Compare.
The function is continuous at the point exactly when the limit exists and equals the defined value.

Worked Examples

Worked Example: A Polynomial at a Point

Determine whether \(f(x,y)=3x^2-2xy+y^2\) is continuous at \((1,2)\).

First evaluate the function at the point:

$$ f(1,2)=3(1)^2-2(1)(2)+(2)^2 =3-4+4 =3. $$

This is a polynomial, so direct substitution in the joint limit gives

\lim_{(x,y)\to(1,2)}f(x,y) =3(1)^2-2(1)(2)+(2)^2 =3-4+4 =3.

The limit exists and equals \(f(1,2)=3\). Therefore \(f\) is continuous at \((1,2)\). As a check, substituting the computed point into the original formula gives \(3\), the same number obtained for the limit.

Worked Example: A Different Value Assigned at the Origin

Consider the function defined on \(\mathbb{R}^2\) by

g(x,y)= \begin{cases} x^2+y^2, & (x,y)\ne(0,0),\\ 4, & (x,y)=(0,0). \end{cases}

Determine whether \(g\) is continuous at \((0,0)\). The value at the point is assigned by the second branch, so

$$ g(0,0)=4. $$

To compute the limit, take any \((x,y)\ne(0,0)\). Then \(g(x,y)=x^2+y^2\). Write \(r=\sqrt{x^2+y^2}\), the distance from \((x,y)\) to the origin. Thus

$$ |g(x,y)-0| =x^2+y^2 =r^2. $$

Given \(\varepsilon>0\), choose \(\delta=\sqrt{\varepsilon}\). Whenever \(0<r<\delta\), we have \(r^2<\delta^2=\varepsilon\). Therefore \(\lim_{(x,y)\to(0,0)}g(x,y)=0\). Since this limit is \(0\), while \(g(0,0)=4\), the two are unequal. Hence \(g\) is not continuous at the origin.

As a further check, the points \((1/n,0)\) approach the origin, and for every \(n\in\mathbb{N}\), \(g(1/n,0)=1/n^2\), which approaches \(0\), not \(4\). The values near the origin therefore do not approach the function’s assigned value there.

Worked Example: Checking a Point Where the Formula Changes

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

h(x,y)= \begin{cases} x+y, & x\le 0,\\ 2x+y, & x>0. \end{cases}

Determine whether \(h\) is continuous at \((0,2)\). Since \(0\le 0\), the first branch applies at the point, giving

$$ h(0,2)=0+2=2. $$

To check the joint limit, set \(r=\sqrt{x^2+(y-2)^2}\), the distance from \((x,y)\) to \((0,2)\). If \(x\le 0\), then

|h(x,y)-2| =|x+y-2| \le |x|+|y-2| \le 2r.

The last inequality holds because both \(|x|\le r\) and \(|y-2|\le r\). If \(x>0\), then

|h(x,y)-2| =|2x+y-2| \le 2|x|+|y-2| \le 3r.

Thus, in either branch, \(|h(x,y)-2|\le 3r\). Given any \(\varepsilon>0\), choose \(\delta=\varepsilon/3\). If \(r<\delta\), then

|h(x,y)-2| \le 3r <3\delta =\varepsilon.

This proves that \(\lim_{(x,y)\to(0,2)}h(x,y)=2\). That limit equals the already computed value \(h(0,2)=2\), so \(h\) is continuous at \((0,2)\). The estimate covers points on both sides of the formula change, and the point itself uses the branch \(x\le 0\).

A Useful Sequential Check

The sequential criterion for multivariable limits gives another way to express the same test. At an accumulation point \((a,b)\), continuity means that for every sequence of domain points \((x_n,y_n)\) approaching \((a,b)\), the values \(f(x_n,y_n)\) approach \(f(a,b)\). This follows by applying the sequential criterion for limits with the candidate limit \(L=f(a,b)\).

Equivalent test: If \((a,b)\) is an accumulation point of \(D\), then \(f\) is continuous at \((a,b)\) if and only if every sequence \((x_n,y_n)\in D\) with \((x_n,y_n)\to(a,b)\) satisfies \(f(x_n,y_n)\to f(a,b)\).

A sequence can disprove continuity by showing that function values fail to approach \(f(a,b)\). A few sequences that do approach the correct value do not, by themselves, prove continuity: the test must hold for every sequence approaching the point. The distance estimate in the third example proves the needed claim for all nearby points at once.

Common Mistake: Finding the Limit but Ignoring the Function Value

A frequent error is to compute a joint limit and then declare continuity without checking the value at the point. For the function \(g\) above, the incorrect conclusion would be: “The limit at the origin is \(0\), so \(g\) is continuous there.” This calculation stops too soon. The function’s actual value is \(g(0,0)=4\), not \(0\). The correct comparison is

\lim_{(x,y)\to(0,0)}g(x,y)=0 \qquad\text{but}\qquad g(0,0)=4.

Since \(0\ne4\), \(g\) is not continuous at the origin. A related mistake occurs with a formula that changes at the point: using a nearby branch to evaluate \(f(a,b)\) can give the wrong value. Always identify which branch includes the point before making the comparison.

Why the Test Matters

Continuity describes agreement between a function’s assigned value and the behavior of nearby values. If the limit exists but differs from the value at the point, the function has a mismatch there, as in the second example. If the limit does not exist, it cannot equal the function value, so the function is not continuous there. If the function is not defined at the point, it also fails the definition of continuity there.

For a function defined by cases, the test remains point-specific: determine the value at the chosen point, then test the joint limit as inputs approach it from every direction in the domain. The one-sided-looking labels in a formula, such as \(x\le 0\) and \(x>0\), do not replace this joint test. Nearby points can approach from many directions, and the limit must describe all of them together.

Practical rule: At the point under investigation, write down \(f(a,b)\) explicitly. Then establish the joint limit separately. Only after both are known can continuity be decided.

Check Your Understanding

For each question, use the definition of continuity at a point rather than relying only on a few sample values.

State the two conditions that must hold for continuity at \((a,b)\) when it is an accumulation point of the domain.
If \(\lim_{(x,y)\to(a,b)} f(x,y)=5\) but \(f(a,b)=2\), is \(f\) continuous there? Explain the comparison.
For \(q(x,y)=x^2+y^2\) when \((x,y)\ne(0,0)\), and \(q(0,0)=0\), what is the limit at the origin, and does it equal the defined value?
In the third worked example, which branch determines \(h(0,2)\), and what distance \(r\) is used to estimate nearby values?
Why does checking several sequences that give the correct limiting value not prove continuity by itself?