Introduction
A rational formula can fail to define a function at a particular input because its denominator is zero there. Sometimes the failure is only a hole: nearby values approach a finite number, and assigning that number at the missing input produces a continuous extension.
Continuity alone does not guarantee differentiability in general. For rational functions, however, the structure of the numerator and denominator provides more information. When the limit at a hole is finite, cancellation reduces the local formula to a function whose difference quotient has a finite limit.
Removable Holes and Continuous Extensions
Let \(P\) and \(Q\) be polynomials, with \(Q\) not the zero polynomial, and consider \(f(x)=P(x)/Q(x)\) wherever \(Q(x)\ne 0\). Suppose \(Q(a)=0\). The formula does not define \(f(a)\), so we ask what happens as \(x\) approaches \(a\) through inputs where \(f\) is defined.
The value \(L\) is the only value that makes this extension continuous at \(a\). The original rational function remains undefined at \(a\); it is the extended function \(F\) that has a value there.
To test differentiability at \(a\), use the definition of the derivative. For the extension, it is
provided this limit exists as a finite number. The value \(h=0\) is excluded from the quotient, and \(h\) must be small enough that \(F(a+h)\) is defined.
Worked Example: Extending a Simple Rational Function
Worked Example: Finding the Derivative at a Hole
For \(x\ne 2\), let \(f(x)=(x^2-4)/(x-2)\). Find the continuous extension at \(x=2\), then calculate its derivative there from the difference quotient.
Factor the numerator and cancel the common factor for \(x\ne 2\):
Thus \(\lim_{x\to 2}f(x)=2+2=4\). Define \(F(2)=4\), and let \(F(x)=f(x)\) for \(x\ne2\). For nonzero \(h\),
The quotient is \(1\) for every sufficiently small nonzero \(h\), so its limit is \(1\), and \(F'(2)=1\).
Check: The extension is \(F(x)=x+2\) on both sides of \(2\), with \(F(2)=4\). Its difference quotient at \(2\) is indeed \(1\), as calculated directly.
Why Finite Limits Give Derivatives
The example reflects a general fact about rational functions. If the rational expression has a finite limit at a denominator zero, its numerator must contain at least as many factors of \(x-a\) as its denominator. After cancellation, the extension is either a rational function with nonzero denominator at \(a\), or a product with a positive power of \(x-a\). Both forms have a finite difference-quotient limit there.
Here is a proof, including the case where the numerator is the zero polynomial. Since a nonzero polynomial has only finitely many roots, we can choose a neighborhood of \(a\) in which \(Q\) has no zero other than \(a\). This makes the rational expression defined throughout the punctured neighborhood.
If \(P\) is identically zero, then \(f(x)=0\) at every defined input near \(a\). Its finite limit is \(0\), and the extension is identically zero throughout the neighborhood. Its difference quotient is \(0/h=0\) for \(h\ne0\), so its derivative at \(a\) is \(0\).
Suppose \(P\) is not identically zero. Because \(P(a)=0\) is necessary for a finite limit when \(Q(a)=0\), write \(P(x)=(x-a)^m p(x)\) and \(Q(x)=(x-a)^n q(x)\), where \(m,n\) are positive integers and \(p(a)\ne0\), \(q(a)\ne0\). The factors \(p\) and \(q\) are polynomials.
For \(x\ne a\), the quotient is \((x-a)^{m-n}p(x)/q(x)\). Since \(p(x)/q(x)\) tends to the nonzero number \(p(a)/q(a)\), a negative power of \(x-a\) would make the absolute value of the quotient unbounded. The finite-limit assumption therefore requires \(m\ge n\).
Let \(k=m-n\ge0\) and \(r(x)=p(x)/q(x)\). The denominator \(q(a)\) is nonzero, so \(r\) is differentiable near \(a\). For \(x\ne a\), \(f(x)=(x-a)^k r(x)\). If \(k=0\), the extension has \(F(a)=r(a)\), and its difference quotient is \([r(a+h)-r(a)]/h\), which tends to \(r'(a)\). If \(k\ge1\), the extension has \(F(a)=0\), and its difference quotient is \(h^{k-1}r(a+h)\). This tends to \(r(a)\) if \(k=1\), and to \(0\) if \(k>1\). Every case has a finite limit, proving differentiability at \(a\).
For clarity, the quotient \(r=p/q\) in the case \(k=0\) has a derivative there because \(q(a)\ne0\). Its first-principles quotient can be written as
As \(h\to0\), the polynomial difference quotients tend to \(p'(a)\) and \(q'(a)\), and the denominator tends to \(q(a)^2\ne0\). Thus the limit is \(\bigl(q(a)p'(a)-p(a)q'(a)\bigr)/q(a)^2\), a finite number.
Worked Example: A Higher-Order Cancellation
Worked Example: A Hole with Zero Derivative
For \(x\ne1\), let \(f(x)=(x-1)^2(x+3)/(x-1)\). Find its continuous extension and its derivative at \(1\).
Cancel one factor for \(x\ne1\):
As \(x\to1\), this expression tends to \(0\). Set \(F(1)=0\). With \(h\ne0\),
Taking the limit as \(h\to0\) gives \(F'(1)=4\).
Check: The extension is \(F(x)=(x-1)(x+3)\), including at \(x=1\). Substitution gives \(F(1)=0\), and its difference quotient at \(1\) is \(h+4\), which tends to \(4\).
Worked Example: Cancellation Leaves a Rational Extension
Worked Example: A Nonlinear Continuous Extension
For \(x\ne1\), let \(f(x)=(x-1)(x^2+1)/\bigl((x-1)(x+2)\bigr)\). Find the extension’s value and derivative at \(1\).
After cancellation, valid for \(x\ne1\), the expression is \((x^2+1)/(x+2)\). Since its denominator is \(3\) at \(x=1\), the limit exists and equals \(2/3\). Define \(F(1)=2/3\). For nonzero \(h\) near zero,
As \(h\to0\), this tends to \(4/9\), so \(F'(1)=4/9\).
Check: The extension value is the canceled expression evaluated at \(1\), namely \((1^2+1)/(1+2)=2/3\). The displayed difference quotient simplifies exactly to \((4+3h)/(9+3h)\), whose limit is \(4/9\).
What the Result Does—and Does Not—Say
The finite-limit condition matters. For example, \(f(x)=(x-2)/(x-2)^2=1/(x-2)\) for \(x\ne2\). Its values do not approach a finite number as \(x\to2\), so there is no continuous extension at \(2\) and no derivative there for an extension of this kind.
Nor does the result say that every continuous function is differentiable. It applies to rational functions at a hole, where polynomial factors control the behavior. A finite limit lets us cancel enough powers of \(x-a\) to get a difference quotient with a finite limit. Other kinds of functions can be continuous at a point and still fail to have a derivative there.
If a value other than the limit is assigned at the hole, the resulting function is discontinuous there and therefore cannot be differentiable there. Thus, for a removable hole, the continuous extension is not just convenient: its value is required for differentiability.
A Common Mistake: Treating Cancellation as a New Value
In the first example, it is tempting to cancel \(x-2\) and conclude that the original function has value \(4\) at \(x=2\). That is incorrect. The original formula still has denominator zero at \(2\), so it is undefined there. Cancellation establishes an equal formula only for \(x\ne2\). To create a function defined at \(2\), one must explicitly define the extension by \(F(2)=4\).
The distinction matters when forming the derivative quotient. The quotient must use the extension’s actual value \(F(2)\), not an assumed value of the original function. Once \(F(2)=4\) is stated, the calculation is valid:
Check Your Understanding
Use cancellation and the difference quotient at the missing input where appropriate.
- For \(x\ne3\), let \(f(x)=(x^2-9)/(x-3)\). What value must the continuous extension assign at \(3\), and what is its derivative there?
- For \(x\ne0\), let \(f(x)=x^3/x\). Find the extension’s value and derivative at \(0\).
- Why does a finite limit rule out the numerator having fewer factors of \(x-a\) than the denominator?
- What is the error in saying that canceling a denominator factor automatically defines the original function at the hole?
- For \(x\ne1\), let \(f(x)=(x-1)/(x-1)^2\). Does it have a finite continuous extension at \(1\)? Explain briefly.
