Continuity at a Boundary
A piecewise-defined function may use one formula above a curve, another below it, and a separate formula on the curve itself. To decide whether the function is continuous at a point of that curve, it is not enough to substitute into just one formula. Points can approach from several parts of the domain, and the function must have the same limiting value along all of them.
For a boundary written as \(y=g(x)\), a convenient way to describe its points is \((t,g(t))\). Suppose the formulas on the two sides are \(F_+(x,y)\) and \(F_-(x,y)\), and the formula on the boundary is \(h(t)\). At a boundary point \(p=(a,g(a))\), continuity requires the function values to approach \(h(a)\) whether the input approaches through the upper region, the lower region, or along the boundary.
What Must Be Checked?
Consider a curve \(\Gamma\) that locally separates the plane into two regions \(R_+\) and \(R_-\). Suppose \(f\) is given by \(F_+\) on \(R_+\), by \(F_-\) on \(R_-\), and by \(h\) on \(\Gamma\). At a point \(p\in\Gamma\), there are three relevant restricted limits: one from each side and one along the curve. Any of these approaches that is possible near \(p\) must give the same limit, and that limit must equal \(f(p)\).
When the boundary formula is continuous along the curve, its restricted limit is simply its value at \(p\). In that common situation, the practical test is to check that the limits of the two regional formulas equal the boundary value. But continuity of the boundary formula should not be silently assumed: irregular boundary values can cause a discontinuity even when both regional limits agree with \(f(p)\).
Why the Test Works
The test follows directly from the definition of a limit. A limit of \(L\) means that for every \(\varepsilon>0\), all domain points sufficiently close to \(p\), other than \(p\) itself, have function values within \(\varepsilon\) of \(L\).
If the full limit is \(L\), then the same \(\varepsilon\)-bound holds when the approaching points are restricted to any one piece. So every restricted limit must also be \(L\).
Suppose each of the finitely many relevant pieces has restricted limit \(L\). For a chosen \(\varepsilon>0\), each piece supplies a distance \(\delta_i>0\) that makes its function values within \(\varepsilon\) of \(L\).
Take \(\delta\) to be the smallest of the finitely many \(\delta_i\). Every domain point within distance \(\delta\) of \(p\) belongs to one of the pieces, so its function value is within \(\varepsilon\) of \(L\). This is exactly the definition of the full limit.
This reasoning also explains why each piece matters. If even one piece has a different restricted limit, points approaching through that piece violate the condition for the full limit. A boundary curve is itself one of the pieces, so its values must be checked too.
Worked Example: Matching Formulas Across a Parabola
Worked Example: Matching Formulas Across a Parabola
Define \(f\) by \[ f(x,y)= \begin{cases} x+y, & y>x^2,\\ x+2x^2-y, & y<x^2,\\ x+x^2, & y=x^2. \end{cases} \] Test continuity at an arbitrary point \(p=(a,a^2)\) on the boundary curve.
The boundary formula gives \[ f(a,a^2)=a+a^2. \] From the upper region, the polynomial formula has limit \[ \lim_{\substack{(x,y)\to(a,a^2)\\y>x^2}}(x+y)=a+a^2. \] From the lower region, its limit is \[ \lim_{\substack{(x,y)\to(a,a^2)\\y<x^2}}(x+2x^2-y) =a+2a^2-a^2=a+a^2. \] Along the boundary, a point has the form \((t,t^2)\), and \[ f(t,t^2)=t+t^2\longrightarrow a+a^2 \quad\text{as }t\to a. \] All three restricted limits equal \(f(a,a^2)\). The continuity test therefore shows that \(f\) is continuous at every point of the parabola.
One can also see how the regional formulas approach the target value by subtracting it. In the upper region, \[ (x+y)-(a+a^2)=(x-a)+(y-a^2)\longrightarrow 0. \] In the lower region, \[ (x+2x^2-y)-(a+a^2) =(x-a)+2(x^2-a^2)-(y-a^2)\longrightarrow 0. \] Each term tends to zero as \((x,y)\to(a,a^2)\). These calculations verify the regional limits directly.
Check: The boundary value at \((a,a^2)\) is \(a+a^2\), and the limits from above, below, and along the curve all equal \(a+a^2\). Hence the full limit equals the function value.
Worked Example: A Mismatch from One Side
Worked Example: A Mismatch from One Side
Now define \[ f(x,y)= \begin{cases} x+y+1, & y>x^2,\\ x+2x^2-y, & y<x^2,\\ x+x^2, & y=x^2. \end{cases} \] Determine whether \(f\) is continuous at \(p=(1,1)\).
The boundary value is \[ f(1,1)=1+1=2. \] From below the curve, the formula tends to \[ \lim_{\substack{(x,y)\to(1,1)\\y<x^2}}(x+2x^2-y) =1+2(1)^2-1=2. \] But from above the curve, it tends to \[ \lim_{\substack{(x,y)\to(1,1)\\y>x^2}}(x+y+1) =1+1+1=3. \] The two regional limits disagree, so the full limit does not exist and \(f\) is not continuous at \((1,1)\).
A sequence makes the failure explicit. For each \(n\in\mathbb N\), let \[ (x_n,y_n)=\left(1,1+\frac{1}{n}\right). \] Then \(y_n> x_n^2=1\), so the upper formula applies. The points approach \((1,1)\), but \[ f(x_n,y_n)=1+\left(1+\frac{1}{n}\right)+1 =3+\frac{1}{n}\longrightarrow 3, \] which is not \(f(1,1)=2\).
Check: Approaching through the lower region gives limiting value \(2\), while the displayed upper-region sequence gives limiting value \(3\). The differing values confirm discontinuity.
Worked Example: Why the Boundary Values Also Matter
Worked Example: Why the Boundary Values Also Matter
Let the boundary again be \(\Gamma=\{(x,y):y=x^2\}\), and define \[ f(x,y)= \begin{cases} 0, & y\ne x^2,\\ 0, & y=x^2\text{ and }x=0,\\ 1, & y=x^2\text{ and }x\ne 0. \end{cases} \] Test continuity at the origin.
The function value is \(f(0,0)=0\). From either adjacent region, the function is identically zero, so both regional limits at \((0,0)\) are \(0\). If only those two limits were checked, they would appear to match the value at the origin.
However, approach along the boundary using the points \[ (x_n,y_n)=\left(\frac{1}{n},\frac{1}{n^2}\right). \] Each point lies on \(y=x^2\), and \(x_n\ne 0\), so \(f(x_n,y_n)=1\) for every \(n\). Since \((x_n,y_n)\to(0,0)\) while the function values remain \(1\), the full limit cannot equal \(f(0,0)=0\). In fact, \(f\) is not continuous at the origin.
Check: The regional limits are \(0\), but the boundary values along the displayed sequence are \(1\). The boundary restriction fails the continuity test, so checking the two sides alone is insufficient.
A Common Mistake
A common error is to compare the two regional limits and stop as soon as they agree. Agreement is necessary, but if the boundary formula is not known to be continuous, it is not sufficient. Example 3 demonstrates the problem: both side limits equal \(0\), and the value at the origin is \(0\), yet values along the boundary approach \(1\).
Another error is to substitute the boundary point into a regional formula and treat that calculation as a limit from the region. A formula evaluated at the point gives a candidate value; it does not by itself control values at nearby points in the region. Compute or justify the restricted limit, then compare it with the assigned boundary value.
Why This Test Is Useful
Piecewise definitions often arise when different formulas describe different parts of a domain. The boundary is where those descriptions meet, so testing continuity there determines whether the pieces fit together without a jump. The test also clarifies exactly what must be established: matching formulas on either side, together with compatible values along the boundary.
For a boundary \(y=g(x)\) with a continuous boundary formula \(h(x)\), the check along the curve is usually immediate: \(h(t)\to h(a)\) as \(t\to a\). Then the main work is comparing the limits of the neighboring formulas with \(h(a)\). If the boundary values are not known to vary continuously, check them independently rather than assuming they do.
Check Your Understanding
For each question, identify which limits or values must be compared.
- For a function defined by separate formulas above and below \(y=g(x)\), what restricted limits must be checked at a point on the curve?
- Why does a mismatch between the two regional limits rule out continuity?
- If both regional limits equal \(f(p)\), what additional check may still be needed?
- In the first worked example, what common value do the two regional limits and the boundary restriction approach at \((a,a^2)\)?
- In the third worked example, why do the regional limits fail to detect the discontinuity at the origin?