Filling a Hole Does Not Guarantee a Derivative
A function may be undefined at one point even though its values nearby approach a finite number. On a graph, this missing point appears as a hole. Assigning the limiting value to the function at that point fills the hole and makes the function continuous there.
Continuity is necessary for differentiability, but it is not sufficient. After filling a hole, we must still check how quickly the function's nearby values change relative to the distance from the filled point. The derivative's difference quotient makes that test precise.
Removable Holes and Their Continuous Fill
Suppose \(g\) is defined for values near \(a\), except possibly at \(a\), and that the two-sided limit \(\lim_{x\to a}g(x)\) exists and is finite. The missing value at \(a\) is called a removable hole. It is removable because assigning the limiting value at \(a\) makes the function continuous there.
The value \(L\) is the only value that makes the function continuous at \(a\). If a different number is assigned there, the limit still equals \(L\), but the function value does not equal the limit.
To decide whether this continuous fill is differentiable at \(a\), use the definition of the derivative. For a function defined on an open interval around \(a\), its derivative at \(a\) exists as a finite real number if the limit below exists as a finite number:
Here \(h\ne 0\), and \(a+h\) must be in the function's domain. Because \(f(a)=L\), the difference quotient for a continuous fill is
This expression measures the change in the function's value from the filled point, divided by the change in input. The limit may exist, fail to exist, or become unbounded. Continuity guarantees only that \(g(a+h)-L\) approaches zero; it does not guarantee that this difference approaches zero at the particular rate needed for the quotient to have a finite limit.
Why Differentiability Requires the Continuous Fill
The derivative definition itself shows why a function cannot be differentiable at a point where it is discontinuous. This fact is useful when a hole has been filled with a value other than its limit.
For \(x\ne a\), write the difference \(f(x)-f(a)\) as the product of \(x-a\) and the difference quotient. If the derivative exists, that quotient approaches a finite number as \(x\to a\). Meanwhile, \(x-a\) approaches zero. Their product therefore approaches zero, which is exactly the condition for continuity at \(a\).
As \(x\to a\), the first factor approaches \(0\), and the second factor approaches \(f'(a)\), a finite number. Thus \(f(x)-f(a)\to 0\), so \(f(x)\to f(a)\). This proves that differentiability implies continuity.
For a function with a removable hole, this result gives a quick first test: if the assigned value does not equal the finite limit, the function is not continuous and therefore cannot be differentiable at the filled point. If the value is chosen correctly, continuity is established, but the difference quotient must still be tested.
A Direct Test at the Filled Point
The following steps keep the two questions separate: did the chosen value make the function continuous, and does the resulting function have a derivative?
Compute \(L=\lim_{x\to a}g(x)\) using the formula valid away from the hole.
For continuity, the function must satisfy \(f(a)=L\). If not, it is discontinuous and not differentiable at \(a\).
When \(f(a)=L\), calculate \(\frac{g(a+h)-L}{h}\) for \(h\ne 0\).
If the two-sided limit as \(h\to 0\) exists and is finite, the filled function is differentiable at \(a\), and that limit is \(f'(a)\).
Worked Examples
Worked Example: Filling a Rational Hole and Finding the Derivative
For \(x\ne 3\), let \(g(x)=\frac{x^2-9}{x-3}\). Fill the hole continuously at \(x=3\), then determine whether the filled function is differentiable there.
For \(x\ne 3\), factor the numerator and cancel the common factor:
Therefore, as \(x\to 3\), \(g(x)\to 3+3=6\). The continuous fill is \(f(3)=6\), and for \(x\ne3\), \(f(x)=x+3\). Since \(\lim_{x\to3}f(x)=6=f(3)\), the filled function is continuous at \(3\).
Now use the difference quotient. For \(h\ne0\), \(f(3+h)=(3+h)+3=6+h\), so
The quotient is \(1\) for every nonzero \(h\) near zero, so its limit is \(1\). Thus \(f'(3)=1\). As a check, substituting \(f(3)=6\) and \(f(3+h)=6+h\) into the quotient gives \(1\), as required by the derivative definition.
Worked Example: A Continuous Fill That Is Not Differentiable
For \(x\ne0\), let \(g(x)=|x|\), and fill the hole at \(0\) with the limiting value. Is the filled function differentiable there?
Since \(|x|\to0\) as \(x\to0\), the continuous assignment is \(f(0)=0\). The function is continuous at \(0\), because its limit and its assigned value are both zero.
For \(h\ne0\), the difference quotient is
If \(h>0\), then \(|h|=h\), so the quotient is \(1\). If \(h<0\), then \(|h|=-h\), so the quotient is \(-1\). The right-hand limit is \(1\), while the left-hand limit is \(-1\). Since they are unequal, the two-sided limit does not exist, and \(f'(0)\) does not exist.
The check is direct: values on the two sides give different quotient limits, so no single finite derivative can satisfy the definition. Filling the hole made the function continuous, but did not make it differentiable.
Worked Example: Choosing the Wrong Value at the Hole
For \(x\ne2\), let \(g(x)=\frac{x^3-8}{x-2}\). Suppose we assign \(f(2)=11\). Determine whether \(f\) is continuous or differentiable at \(2\).
Factor the numerator using the difference of cubes:
Thus, for \(x\ne2\), \(g(x)=x^2+2x+4\), and
The assigned value \(f(2)=11\) is not \(12\), so \(f\) is discontinuous at \(2\). To see the failure in the derivative quotient, for \(h\ne0\) we have
As \(h\to0\), the term \(1/h\) is unbounded, so the quotient has no finite two-sided limit. This confirms that the function is not differentiable at \(2\). The continuous fill would have been \(f(2)=12\), not \(11\).
A Common Mistake: Stopping After Cancellation
A common error is to cancel a factor, obtain a simpler expression, and then treat that expression as if it had already settled the derivative question at the missing point. Cancellation is valid for \(x\ne a\); it does not assign a value to the original function at \(a\), nor does it by itself establish the derivative of a chosen fill.
For example, from \(\frac{x^2-9}{x-3}=x+3\) for \(x\ne3\), it would be incorrect to claim that the original function automatically has derivative \(1\) at \(3\). The original function is undefined there. The correction is to assign \(f(3)=6\), check that this gives continuity, and then evaluate the quotient:
Only after this limit calculation can we conclude that the filled function is differentiable at \(3\) and that its derivative is \(1\).
What the Test Tells You
At a removable hole, the limiting value identifies the only continuous fill. The derivative test then asks a finer question: does the change in function value, divided by the change in input, settle to a finite number? The examples show three distinct outcomes: the correct fill can have a finite derivative; it can be continuous but fail the derivative test; or an incorrect fill can remain discontinuous and therefore cannot be differentiable.
Keep these conclusions separate. First compare the assigned value with the limit. Then, if they match, compute the difference quotient at the filled point. Do not infer differentiability from cancellation, a graph that appears smooth, or continuity alone.
Check Your Understanding
For each question, distinguish the value needed for continuity from the separate test for differentiability.
- For \(x\ne1\), let \(g(x)=\frac{x^2-1}{x-1}\). What value fills the hole continuously at \(x=1\), and what is the derivative of the filled function there?
- For \(x\ne0\), let \(g(x)=|x|\), with \(f(0)=0\). What are the difference-quotient limits from the right and left?
- For \(x\ne2\), let \(g(x)=\frac{x^2-4}{x-2}\), but assign \(f(2)=5\). Is \(f\) continuous at \(2\)? Explain using the limit.
- Why does a finite derivative at a point imply continuity there?
- After correctly filling a removable hole, what limit must be evaluated to test differentiability?
