When a Function Has a Hole
A function can approach a definite, finite value as its input approaches a point, even when the function is not defined at that point. It can also be defined there but assigned a value that does not match what nearby function values approach. In either case, the problem is confined to one point: assigning the appropriate value there makes the function continuous.
This kind of break is called a removable discontinuity. The key test is whether the two-sided limit exists as a finite real number. If it does, that limit tells us exactly which value is needed to fill the hole or correct the mismatched value.
The phrase “sufficiently close to \(a\), except possibly at \(a\)” means that there is some interval around \(a\) where the function is defined at every point other than \(a\). The limit \(L\) describes the values of the function near \(a\); it does not depend on what value, if any, the function has at \(a\).
Finding the Value That Fills the Hole
For a function to be continuous at \(a\), its value at \(a\) must agree with the limit of its values as \(x\) approaches \(a\). Thus, once the finite limit \(L\) is known, the required value is \(f(a)=L\). If the function was undefined at \(a\), we can add that value to define an extension. If it was already defined there with a different value, we must change that value to make a continuous function.
These two situations are related, but the terminology is worth keeping precise. A continuous extension fills in a point omitted from the function’s domain while agreeing with the original function everywhere on that domain. If the original function already has a wrong value at \(a\), changing it does not preserve the original function at every point of its domain; it is a correction of that function’s value.
Worked Example: Filling a Hole After Factoring
Let \(f(x)=\frac{x^2-9}{x-3}\) for \(x\ne 3\). Find the value to assign at \(x=3\) to make the extension continuous.
First, factor the numerator. For \(x\ne 3\), the denominator is nonzero, so cancellation is valid:
We can now find the limit using the simplified expression, which agrees with the original function for all nearby \(x\ne 3\):
The original formula has no value at \(x=3\), because its denominator would be \(3-3=0\). Define \(F\) on all real numbers by retaining the original values away from \(3\) and setting \(F(3)=6\). Then \(F(3)=6\), and \(\lim_{x\to 3}F(x)=6\), so \(F\) is continuous at \(3\). As a check, \(F(x)=x+3\) for every real \(x\), including \(x=3\), and this formula gives \(F(3)=3+3=6\).
Why the Limit Gives the Correct Value
Continuity at \(a\) means that \(f(a)\) is defined and \(\lim_{x\to a}f(x)=f(a)\). Suppose instead that we start with a function \(f\) defined near \(a\) except at \(a\), and that \(\lim_{x\to a}f(x)=L\) is finite. Define a new function \(F\) by keeping the old values away from \(a\) and setting \(F(a)=L\). The values of \(F\) near \(a\), for \(x\ne a\), are exactly the old values of \(f\), so their limit remains \(L\). This proves that \(F\) is continuous at \(a\).
The same reasoning explains why there is only one possible value that works. If we set \(F(a)=c\) and want \(F\) to be continuous at \(a\), the definition of continuity requires \(\lim_{x\to a}F(x)=F(a)\). The limit on the left is still \(L\), because changing a value at the single point \(a\) does not change the values at nearby points other than \(a\). Therefore \(c=L\). Any other choice leaves the function discontinuous.
This result relies on the existence of a finite two-sided limit. A value cannot repair a break if the nearby function values do not approach one finite number.
When a Value Is Present but Wrong
A function may already be defined at \(a\), but its assigned value may differ from the limit. The limit must be calculated from nearby values; simply substituting \(a\) into the stated function value does not determine the limit. To make the function continuous, retain its values away from \(a\) and replace its value at \(a\) with the limit.
Worked Example: Correcting a Mismatched Value
Define \(f(x)=\frac{x^2-4}{x-2}\) for \(x\ne 2\), and set \(f(2)=1\). Find the value at \(2\) that makes the function continuous.
For \(x\ne 2\), factor and cancel:
Thus the limit is
The given value is \(f(2)=1\), which differs from the limit \(4\). Therefore the function is not continuous at \(2\). Replace the assigned value by defining \(F(2)=4\), and let \(F(x)=f(x)\) for \(x\ne 2\). Now \(F(2)=4=\lim_{x\to 2}F(x)\), so \(F\) is continuous at \(2\). Checking the simplified formula, \(F(x)=x+2\) for every \(x\), and \(F(2)=2+2=4\), as required.
Checking Both Sides of a Piecewise Formula
For a piecewise function, the limit must agree from the left and from the right. A finite limit exists only when both one-sided limits exist and are equal. When they have the same value, that common value is the one to use at the missing point. This check matters because a formula on one side alone cannot determine the two-sided limit.
Worked Example: Joining Two Formulas at a Missing Point
Let
The function is not defined at \(x=1\). Determine whether the missing value can be filled to make it continuous.
From the left, use the first formula:
From the right, use the second formula:
Since both one-sided limits equal \(2\), the finite two-sided limit exists and equals \(2\). Define \(F(1)=2\), keeping the given formulas for \(x<1\) and \(x>1\). Then \(\lim_{x\to 1}F(x)=2=F(1)\), so \(F\) is continuous at \(1\). The left-side check gives \(1+1=2\); the right-side check gives \(3(1)-1=2\). Both agree with the assigned value.
A Common Mistake: Cancelling Does Not Change the Original Domain
When a factor cancels, the simplified formula can help evaluate a limit, but it does not automatically redefine the original function at the point that made the denominator zero. For example, \(\frac{x^2-9}{x-3}\) and \(x+3\) have equal values whenever \(x\ne 3\), but the original fraction is still undefined at \(3\). The simplified expression represents a continuous extension only if we explicitly define its value there.
Another error is to focus only on the value printed in a piecewise definition. Continuity depends on both that value and the limit from nearby inputs. A function can have the correct limit but the wrong assigned value, or have a chosen value that seems plausible while the two-sided limit does not exist. Check the limit first, then compare or assign the function value.
How to Test for a Removable Discontinuity
Use the following sequence when a function appears to have a hole or a suspicious value at \(a\):
Confirm that the function is defined on both sides of \(a\) sufficiently close to \(a\), even if it is not defined at \(a\) itself.
Use the formulas valid for nearby inputs. For a piecewise function, calculate the left-hand and right-hand limits separately and compare them.
If both sides approach the same real number \(L\), the limit is finite and the point can be repaired by assigning the value \(L\).
If the function is missing at \(a\), define its value to be \(L\). If a value is already present, replace it with \(L\) when making a continuous version.
Check that the resulting function has value \(L\) at \(a\) and limit \(L\) as \(x\to a\).
Why Removable Discontinuities Matter
Identifying a removable discontinuity separates the behavior of a function near a point from the value assigned exactly at that point. This distinction is useful when simplifying expressions, interpreting graphs, and deciding whether a function can be made continuous by changing one value. It also prevents a common logical slip: a simplified formula can describe the nearby behavior without describing the original function’s domain at the excluded point.
The essential test is direct: there must be one finite two-sided limit. When that limit is \(L\), the only value that makes the function continuous at \(a\) is \(L\). The function might originally omit \(a\), or it might assign a different value there; in either case, the limit identifies the necessary repair.
Check Your Understanding
For each question, distinguish the function’s value at the point from its nearby behavior.
- For \(f(x)=\frac{x^2-16}{x-4}\) when \(x\ne 4\), what value fills the hole at \(x=4\)?
- If \(\lim_{x\to a}f(x)=5\) but \(f(a)=0\), what value should replace \(f(a)\) to make the function continuous?
- For a piecewise function, the left-hand limit at \(a\) is \(3\) and the right-hand limit is \(3\). What value should be assigned at \(a\) to make it continuous?
- Why does cancelling a common factor not, by itself, assign a value to the original function at the point where the factor cancels?
- Why is there only one value that can be assigned at a missing point to make the function continuous?
