Why a Hole Must Be Repaired Before Applying IVT
A function can behave smoothly on both sides of a point and still fail to be continuous there because its value is missing or incorrect. The Intermediate Value Theorem cannot be applied to a function that is discontinuous on the interval in question. If the discontinuity is removable, however, assigning the appropriate limiting value can produce a continuous function. We can then check whether that repaired function satisfies the theorem’s other conditions.
There are two distinct questions to keep in view. First, does the repaired function meet the hypotheses of the Intermediate Value Theorem on the chosen closed interval? Second, does the conclusion about the repaired function also tell us something about the original function? The second question matters especially if the original function was defined at the repaired point with a different value.
What Makes a Discontinuity Removable?
Suppose \(f\) is defined at points near \(c\), though it may or may not be defined at \(c\). If the finite limit \(\lim_{x\to c}f(x)=L\) exists, then the function has a well-defined value toward which it approaches at \(c\). A discontinuity at \(c\) is called removable when the only problem at that point is that \(f(c)\) is missing or differs from \(L\).
The qualification about restriction is important. A continuous extension of a function must agree with that function everywhere on its domain. If the original function already has a value at \(c\) that differs from the limit, changing that value does not give an extension of the original function on its full domain. It gives an extension of the restriction that omits \(c\), and it is a repaired version of the original function.
For instance, if \(\lim_{x\to c} f(x)=4\) but \(f(c)=10\), assigning \(4\) at \(c\) changes the original function’s value there. The new function can be continuous, but it does not agree with the original function at \(c\).
Why Assigning the Limit Gives Continuity
Here is the reason the repair works. Assume that \(f\) is defined on a punctured neighborhood of \(c\) and that \(\lim_{x\to c}f(x)=L\). Define \(F\) to agree with \(f\) at nearby points other than \(c\), and set \(F(c)=L\). By the definition of the limit, for every \(\varepsilon>0\), there is a \(\delta>0\) such that
For \(x\ne c\), \(F(x)=f(x)\), while \(F(c)=L\). Therefore the same implication gives \(|F(x)-F(c)|<\varepsilon\) whenever \(0<|x-c|<\delta\). At \(x=c\), the difference is exactly zero. Thus \(|x-c|<\delta\) implies \(|F(x)-F(c)|<\varepsilon\), which is the definition of continuity of \(F\) at \(c\).
This argument establishes continuity at the repaired point. It does not by itself establish that \(F\) is continuous at every other point of an interval. Those points must also be checked before using IVT.
The Conditions for Using IVT After a Repair
The Intermediate Value Theorem says that if a function \(F\) is continuous on a closed interval \([a,b]\), and a target \(T\) lies between \(F(a)\) and \(F(b)\), then there is some \(d\in[a,b]\) such that \(F(d)=T\). If \(T\) is strictly between the endpoint values, then \(d\) can be taken in \((a,b)\). If \(T\) equals an endpoint value, that endpoint itself already gives the conclusion.
When a removable discontinuity is present, the theorem applies to the repaired function only after all these checks succeed:
Verify that the limit at the missing or mismatched point exists and is finite.
Assign the limit at the point and be precise about whether this extends a function with a hole or replaces an existing value.
Confirm that the repaired function is defined and continuous at every point of the chosen closed interval, including its endpoints.
Calculate the two endpoint values and verify that the stated target lies between them.
The closed interval must be part of the repaired function’s domain. Continuity at the repaired point alone is not enough if there is another discontinuity elsewhere in \([a,b]\), or if the function is not defined at an endpoint.
Worked Examples
Worked Example: A Hole Inside the Interval
For \(x\ne 2\), let \(f(x)=\dfrac{x^2-4}{x-2}\). Repair the hole and determine whether the repaired function must take the target value \(9/2\) on \([1,3]\).
First factor the numerator. For \(x\ne 2\),
Consequently,
The original function has no value at \(2\). Define \(F(2)=4\) and \(F(x)=f(x)\) for \(x\ne2\). Then \(F(x)=x+2\) on all of \([1,3]\), including at \(2\), so \(F\) is continuous there.
Now check the endpoints and the target:
All the hypotheses hold: \(F\) is continuous on \([1,3]\), and \(9/2\) lies between its endpoint values. IVT therefore guarantees some \(d\in[1,3]\) for which \(F(d)=9/2\). As a check, solving \(d+2=9/2\) gives \(d=5/2\), which lies in \([1,3]\), and \(F(5/2)=5/2+2=9/2\). Since \(5/2\ne2\), the original function is also defined there and takes the target value.
Worked Example: Correcting an Existing Value
Define \(f(x)=\dfrac{x^2-1}{x-1}\) for \(x\ne1\), and set \(f(1)=9\). Repair the discontinuity and decide whether the repaired function takes the value \(5/2\) on \([0,2]\).
For \(x\ne1\), factor and cancel:
Thus \(\lim_{x\to1}f(x)=1+1=2\), whereas the originally assigned value is \(f(1)=9\). The function is discontinuous at \(1\). To make the domain distinction explicit, restrict the original function to its domain with \(1\) removed, and extend that restriction by setting \(F(1)=2\). The resulting repaired function is \(F(x)=x+1\) on \([0,2]\). It is not an extension of the original function on its full domain, because \(F(1)=2\ne9=f(1)\).
Check continuity and endpoint values:
The target is between these values because \(1<5/2<3\). IVT guarantees a \(d\in[0,2]\) with \(F(d)=5/2\). Directly, \(d+1=5/2\) gives \(d=3/2\), and \(F(3/2)=3/2+1=5/2\). This witness is not the repaired point, so the original function also takes the target there: \(f(3/2)=5/2\).
A tempting but incorrect statement would be that the repaired function agrees with the original everywhere the original was defined. At \(x=1\), that would require \(2=9\), which is false. The repair agrees with the original only where \(x\ne1\).
Worked Example: A Repair Does Not Guarantee Every Target
For \(x\ne3\), let \(g(x)=\dfrac{x^2-9}{x-3}\). Fill the hole, then determine whether IVT guarantees that the repaired function takes the target value \(2\) on \([0,2]\).
For \(x\ne3\),
Hence \(\lim_{x\to3}g(x)=6\). Assigning \(G(3)=6\) repairs the hole, and \(G(x)=x+3\) is continuous on its domain. In particular, it is continuous on \([0,2]\); the repaired point \(3\) is outside this interval, but the formula already defines \(G\) throughout \([0,2]\).
The endpoint values are
The target \(2\) is not between \(3\) and \(5\), so the endpoint condition for IVT is not satisfied. IVT gives no conclusion that the target occurs. In fact, \(G(x)=x+3\ge3\) for every \(x\in[0,2]\), so \(G(x)=2\) has no solution in that interval. The repair supplies continuity; it does not make every target occur.
What IVT Says About the Original Function
If the original function was undefined at \(c\), the repaired function’s value at \(c\) is new. IVT guarantees a point where the repaired function takes the target, but that point might be \(c\). If the target equals the limiting value \(L\), it is possible that the repaired function takes the target only at \(c\), where the original function has no value.
For example, in the first worked example, the repaired function takes the value \(4\) at \(x=2\), but the original function is undefined there. In contrast, the target \(9/2\) had a witness \(x=5/2\), where the original function was defined. When the original function had a mismatched value at \(c\), changing it can also alter whether the original function takes a target at that point.
A Common Mistake
A frequent error is to notice a finite limit, fill the hole, and immediately claim that IVT applies. This skips two independent checks: continuity at every other point of the interval and whether the target lies between the endpoint values. Another error is to treat a changed point value as though the repaired function still agreed with the original everywhere.
A careful argument names the function to which IVT is being applied, identifies the interval, and displays both endpoint values. It then states exactly which function the conclusion concerns. These steps keep the repair and the theorem’s hypotheses separate.
Check Your Understanding
For each question, distinguish the repaired function from the original wherever their values may differ.
- If \(\lim_{x\to c}f(x)=L\) and \(f(c)\) is undefined, what value should be assigned to repair the discontinuity?
- If \(f(c)=7\) but \(\lim_{x\to c}f(x)=2\), does assigning \(2\) produce an extension of \(f\) on its entire original domain? Explain.
- A repaired function \(F\) is continuous on \([a,b]\), with \(F(a)=1\) and \(F(b)=6\). Which target values are guaranteed by IVT?
- Why does continuity of a repaired function at the filled point alone not establish continuity on a whole interval?
- If IVT guarantees \(F(d)=L\) and \(d\) is the point that was missing from the original domain, what additional issue must be checked before claiming the original function takes the value \(L\)?
