Removable Discontinuities

A Discontinuity That Can Be Repaired

A discontinuity can arise because nearby outputs fail to settle down, as in the oscillatory or unbounded behaviors considered when studying discontinuous functions. But there is a simpler possibility: the outputs near a point approach one finite number, while the function has a different value at the point. In that case, changing just the value at the point makes the function continuous there. This tutorial identifies precisely when that repair is possible and why its value is forced.

The relevant limit concerns inputs in the domain other than the point itself. This distinction matters: the value \(f(a)\) has no role in determining the limit as \(x\) approaches \(a\), but it does matter when deciding whether \(f\) is continuous at \(a\). As in earlier discussions of continuity and limits, approaches are always restricted to points in the function’s domain.

The condition \(L\ne f(a)\) is essential. If the limit exists and equals \(f(a)\), the Continuity as a Limit theorem says that \(f\) is continuous at \(a\), so there is no discontinuity to remove. If the limit does not exist as a finite real number, changing only \(f(a)\) cannot make \(f\) continuous there.

The Unique Value That Repairs Continuity

Suppose \(f\) has a finite limit \(L\) through the punctured domain near \(a\). Define a new function by leaving every value except the value at \(a\) unchanged, and assigning \(L\) at \(a\). The resulting function is continuous at \(a\). Moreover, no other assignment can make a function that agrees with \(f\) at nearby points continuous there.

Proof. To prove continuity of \(F\) at \(a\), fix \(\varepsilon>0\). By the definition of the punctured limit, there is a \(\delta>0\) such that whenever \(x\in E\), \(0<|x-a|<\delta\), we have \(|f(x)-L|<\varepsilon\). For such \(x\), \(F(x)=f(x)\) and \(F(a)=L\), so \(|F(x)-F(a)|<\varepsilon\). If \(x=a\), then \(|F(x)-F(a)|=0<\varepsilon\). Thus the epsilon-delta definition gives continuity of \(F\) at \(a\).

For uniqueness, since \(a\) is an accumulation point of \(E\), for each positive integer \(n\) we can choose \(x_n\in E\setminus\{a\}\) with \(|x_n-a|<1/n\). Then \(x_n\to a\). The assumed punctured limit gives \(f(x_n)\to L\), and \(G(x_n)=f(x_n)\) for every \(n\). Since \(G\) is continuous at \(a\), the Sequential Criterion for Continuity gives \(G(x_n)\to G(a)\). The same sequence therefore has limits \(L\) and \(G(a)\); uniqueness of limits implies \(G(a)=L\). \(\square\)

This theorem explains both parts of the word “removable.” The discontinuity can be removed by changing one function value, and the value required for the repair is unique. The limit itself is determined by all nearby domain values; it is not a matter of choosing a convenient replacement.

Worked Examples: Finding the Limiting Value

What a Removable Discontinuity Is Not

The defining feature is a finite limiting value through the punctured domain. If nearby values become unbounded, or if they approach different values along different sequences, a single change at \(a\) cannot make the function continuous there. For example, defining \(q(0)=0\) and \(q(x)=1/x\) for \(x\ne0\) does not give a removable discontinuity at \(0\): along \(x_n=1/n\), the outputs are \(q(x_n)=n\), which do not converge to any finite real number.

A common source of mistakes is to treat a formula as though it determines the function value even where that formula is undefined. In the first worked example, the simplified expression \(x+2\) correctly describes \(f(x)\) for \(x\ne2\), but it does not describe \(f(2)\). The limit is \(4\); the originally assigned value is \(0\). A removable discontinuity occurs precisely because these two numbers differ.

There is also a distinction between a discontinuity and a missing point in the domain. Under the definition used here, discontinuity at \(a\) refers to a function defined at \(a\). If a formula is defined only on \(E\setminus\{a\}\), then \(a\) is not a point where that function is discontinuous. When the punctured limit exists finitely, one can instead ask whether the function has a continuous extension to \(E\) by assigning a value at \(a\). The Unique Continuous Repair theorem shows that this extension, if made to agree with the original function elsewhere, has only one possible value at \(a\).

Adding a Continuous Function

The limiting behavior responsible for a removable discontinuity is preserved in a useful way by addition. A function continuous at \(a\) has nearby values approaching its value there. Adding those values to a function with a finite punctured limit shifts the limit by exactly that continuous value.

Proof. Fix \(\varepsilon>0\). The punctured limit of \(f\) gives a \(\delta_1>0\) such that \(x\in E\) and \(0<|x-a|<\delta_1\) imply \(|f(x)-L|<\varepsilon/2\). Continuity of \(h\) at \(a\) gives a \(\delta_2>0\) such that \(x\in E\) and \(|x-a|<\delta_2\) imply \(|h(x)-h(a)|<\varepsilon/2\). Set \(\delta=\min\{\delta_1,\delta_2\}\). For \(x\in E\) with \(0<|x-a|<\delta\), the triangle inequality yields \[ |(f(x)+h(x))-(L+h(a))| \leq |f(x)-L|+|h(x)-h(a)| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. \] This proves the stated punctured limit. At the point itself, \((f+h)(a)=f(a)+h(a)\). If \(f(a)\ne L\), then \(f(a)+h(a)\ne L+h(a)\), so the function value differs from its finite punctured limit. Thus \(f+h\) has a removable discontinuity at \(a\). \(\square\)

This result can save work when a complicated expression consists of a part with a removable discontinuity plus a part already known to be continuous. It does not say that every operation preserves removability without conditions; the conclusion here uses the continuity of \(h\) at the same point and the finite punctured limit of \(f\).