Why Redefining One Function Value Can Help
A graph may have a single hole even when the function values on either side of that hole approach one clear, finite height. In that situation, changing the function at just the hole can remove the discontinuity without changing the behavior nearby.
The key is not to guess a value from the appearance of a formula. Find the limit as \(x\) approaches the point, then assign the function that exact limiting value there. The continuity test confirms whether the repair works.
What Counts as a Removable Discontinuity?
Let \(a\) be an interior point of the function's domain or of the interval where we want to define an extension. A discontinuity at \(a\) is removable when the function approaches a finite value \(L\) as \(x\) approaches \(a\), but the function is either not defined at \(a\), or its value at \(a\) is not \(L\).
In other words, the nearby behavior already has a well-defined height. The problem lies only in the value assigned—or not assigned—at the point itself. To repair the discontinuity, leave the function unchanged for \(x\ne a\) and set its value at \(a\) equal to \(L\).
A punctured interval around \(a\) means values close to \(a\), but excluding \(a\) itself. The two-sided limit must exist: the left-hand and right-hand limits must agree and be finite. If the function already has a value at \(a\), that value is replaced by \(L\) if necessary.
Why the Repair Works
Continuity at \(a\) has three requirements: the function is defined at \(a\), the limit as \(x\) approaches \(a\) exists, and that limit equals the function's value at \(a\). The proposed repair ensures all three conditions.
For every \(x\ne a\) near \(a\), set \(g(x)=f(x)\). The limit is therefore unchanged: \(\lim_{x\to a}g(x)=\lim_{x\to a}f(x)=L\).
Set \(g(a)=L\). This ensures that \(g\) is defined at \(a\).
Because \(\lim_{x\to a}g(x)=L=g(a)\), the limit equals the function value. Thus \(g\) is continuous at \(a\).
This also shows that the repair is unique. If an extension changes only the value at \(a\) and is continuous there, continuity requires its value at \(a\) to equal the existing limit \(L\). No other assigned value can make it continuous.
Worked Examples
Worked Example: Repair a Hole in a Rational Expression
Suppose \(f(x)=\frac{x^2-9}{x-3}\) for \(x\ne3\), and \(f(3)\) is currently undefined. Find the value that repairs the discontinuity.
First, factor the numerator:
For \(x\ne3\), the denominator is nonzero, so cancellation gives \(f(x)=x+3\). This simplified expression agrees with the original function at every nearby input other than \(3\). Therefore:
Define \(g(x)=f(x)\) when \(x\ne3\), and set \(g(3)=6\). Check the continuity conditions: \(g(3)=6\), the limit exists and equals \(6\), and thus \(\lim_{x\to3}g(x)=6=g(3)\). The repaired function is continuous at \(3\).
As a direct check, for all \(x\), the repaired function has the same values as \(x+3\): away from \(3\) by cancellation, and at \(3\) because \(g(3)=6=3+3\). The limit and assigned value agree.
Worked Example: Replace an Incorrect Existing Value
Consider the function
Find the value that should replace \(-4\) to make \(f\) continuous at \(2\).
Factor the numerator and simplify for \(x\ne2\):
The limit is therefore
The current value is \(f(2)=-4\), which is not \(5\). To repair the discontinuity, redefine only the point value: set \(g(2)=5\), with \(g(x)=f(x)\) for \(x\ne2\). Then \(\lim_{x\to2}g(x)=5\) and \(g(2)=5\), so \(\lim_{x\to2}g(x)=g(2)\). The repaired function is continuous at \(2\).
Check the replacement: the simplified nearby expression approaches \(5\), and the new point value is exactly \(5\). Leaving the old value \(-4\) in place would fail the continuity condition.
Worked Example: Find a Limit Before Assigning the Value
For \(x\ne4\), let \(f(x)=\frac{\sqrt{x+5}-3}{x-4}\), and suppose \(f(4)\) is undefined. Find a continuous extension at \(4\).
Direct substitution into the displayed fraction gives a numerator and denominator both equal to zero, so it does not determine the limit. Multiply by the conjugate of the numerator, \(\sqrt{x+5}+3\):
The numerator simplifies to \(x-4\). For \(x\ne4\), cancel that factor:
Now evaluate the limit using the simplified expression:
Define \(g(4)=\frac16\), keeping \(g(x)=f(x)\) for \(x\ne4\). The limit remains \(\frac16\), and the assigned value is also \(\frac16\). Thus \(\lim_{x\to4}g(x)=g(4)\), so the extension is continuous at \(4\).
As a final check, the simplified expression evaluated at \(4\) is \(\frac{1}{\sqrt9+3}=\frac16\), exactly the value assigned to the extension.
What a Point-Value Change Cannot Fix
Redefining a function at one point cannot change its nearby behavior. In particular, it cannot create a finite two-sided limit when the left- and right-hand limits differ, when the function grows without bound, or when the function has no limit because it continues to oscillate. Those cases do not satisfy the condition needed for this repair.
For example, if the left-hand limit at \(a\) is \(2\) and the right-hand limit is \(5\), then no single value assigned at \(a\) can make both sides approach that value. If the function tends to \(+\infty\), no finite point value can equal that infinite behavior. The repair described here applies specifically when the two-sided limit exists and is finite.
Common Mistake: Treating Cancellation as a Point Value
In the first example, cancelling \(x-3\) gives \(x+3\), but the cancellation is valid only for \(x\ne3\). A mistaken calculation might say:
The conclusion happens to give the correct repaired value, but the equality for every \(x\) is false: the original fraction is undefined at \(x=3\), so it has no value there. Cancellation computes the expression's values away from the hole; the limit then determines what value to assign at the hole.
The correct reasoning keeps these steps separate:
Another error is to assign \(0\) just because direct substitution produces \(0/0\). That form does not give the limiting value. Simplify or otherwise evaluate the limit first, then assign the resulting finite number.
A Reliable Repair Procedure
When a function appears to have a removable discontinuity at \(a\), use this order:
- Check that the function is defined for inputs close to \(a\), except possibly at \(a\).
- Evaluate the two-sided limit \(\lim_{x\to a}f(x)\), using algebra or a relevant limit result when direct substitution is inconclusive.
- Confirm that the limit exists and is finite. If it does not, a one-point redefinition cannot produce continuity.
- Call the limit \(L\), and define the new point value to be \(g(a)=L\).
- Verify that \(\lim_{x\to a}g(x)=g(a)\). Because the nearby values are unchanged, the limit remains \(L\).
This method matters because it distinguishes a genuine repair from a guess. The limit describes what the function's nearby values require; continuity requires the value at the point to match that limit. Once the limit is known, the correct assignment is forced.
Check Your Understanding
For each question, focus on the limiting value and the continuity condition.
- If \(\lim_{x\to a}f(x)=7\) and \(f(a)\) is undefined, what value should be assigned at \(a\) to obtain continuity?
- For \(x\ne2\), let \(f(x)=\frac{x^2-4}{x-2}\). What point value makes the extension continuous at \(2\)? Show how you found the limit.
- If a function has left-hand limit \(3\) and right-hand limit \(8\) at \(a\), can changing only its value at \(a\) make it continuous? Explain.
- A function has a finite limit \(L\) at \(a\), but its existing value \(f(a)\) is different from \(L\). What must change to repair the discontinuity?
- Why does cancelling a factor in a formula for \(x\ne a\) not, by itself, define the original function's value at \(a\)?
