Canceling Common Factors Near a Hole

Why a Quotient Can Have a Limit Where It Is Undefined

A rational expression may be undefined at a particular input even though its values near that input settle toward a definite number. This happens, for example, when both numerator and denominator contain a factor that becomes zero at the target. The original quotient then has the form zero divided by zero at that input, but the shared factor can be canceled at nearby inputs.

The crucial point is that cancellation simplifies the expression only where the canceled factor is nonzero. A limit concerns values near the target, not necessarily the value at the target itself. Keeping those two ideas separate lets us simplify correctly without claiming that the original function is defined where it is not.

Common Factors and the Domain

Suppose a quotient has the form \(\frac{(x-a)u(x)}{(x-a)v(x)}\). At \(x=a\), the common factor \(x-a\) is zero, so the displayed quotient is undefined there. At any \(x\ne a\) for which the quotient is defined, the factor \(x-a\) is nonzero and can be canceled:

This equality is not an equality at \(x=a\): the left-hand expression is undefined there, and cancellation does not change that fact. But if the simplified expression has a limit as \(x\) approaches \(a\), then it can be used to find the limit of the original expression, provided the two expressions agree at all sufficiently nearby domain points other than \(a\).

The condition \(0<|x-a|\) excludes the target itself. The accumulation-point requirement ensures that there are domain values arbitrarily close to \(a\), so the limit condition is not true merely because there are no nearby inputs to check. The function may be undefined at \(a\), or it may have a value there; either way, that value is not part of the limit condition.

A Justification for Canceling Near the Target

For rational functions, the cancellation method follows from the limit laws and continuity of polynomials. The following result gives its conditions explicitly. Continuity at \(a\) means that the function is defined at \(a\) and that its limit as \(x\) approaches \(a\) equals its value there.

Here is why the hypotheses make the argument work. Since \(v\) is continuous at \(a\) and \(v(a)\ne0\), there is a neighborhood of \(a\) in which \(v(x)\ne0\). In fact, continuity gives a neighborhood where \(|v(x)-v(a)|<|v(a)|/2\), and hence \(|v(x)|>|v(a)|/2>0\). For \(x\ne a\) in that neighborhood, both \(x-a\) and \(v(x)\) are nonzero, so cancellation gives \(f(x)=u(x)/v(x)\).

As \(x\) approaches \(a\), continuity gives \(u(x)\to u(a)\) and \(v(x)\to v(a)\). The quotient law applies because the denominator limit \(v(a)\) is nonzero. Therefore \(\frac{u(x)}{v(x)}\to\frac{u(a)}{v(a)}\). Since the original quotient agrees with this simplified quotient at nearby domain points other than \(a\), the original quotient has the same limit. This proves the result.

In the examples below, polynomial factors play the roles of \(u\) and \(v\). Polynomials are continuous at every real number, so after canceling the shared factor, direct substitution is justified when the remaining denominator is nonzero at the target.

Worked Examples

The Limit Is Not the Function Value

The examples all have an original quotient whose denominator vanishes at the target. Consequently, the original function is not defined there. The limit answers a different question: what number do the function values approach as the input approaches the target through points in the domain?

For example, in the first worked example, the quotient agrees with \(x-3\) for every \(x\ne2\), and those nearby values approach \(-1\). One may define a new function by assigning the value \(-1\) at \(x=2\), but that is an extension of the original function, not a value already supplied by its original quotient. More generally, changing or assigning a function's value at the target does not affect its limit there, as long as its values at nearby points are unchanged.

A Common Mistake

Consider \(f(x)=\frac{x^2-4x+4}{x-2}\), defined only where its denominator is nonzero. A mistaken calculation might say:

The cancellation step is valid only for \(x\ne2\). At \(x=2\), the original denominator is zero, so \(f(2)\) is undefined. The corrected conclusion is about the limit:

The number \(0\) is the limit, not the value of the original function. The same distinction applies even when a graph or a table makes the nearby behavior look continuous through the missing point.

When Cancellation Is Not Enough

After canceling the common factor, inspect the remaining denominator before substituting. The simple direct-substitution conclusion is justified when that denominator is nonzero at the target. If it is still zero there, this cancellation has not yet reduced the expression to a quotient with a nonzero denominator at the target; another valid simplification or method would be needed.

Also check that the cancellation is of a genuine common factor, not merely of terms that appear in a sum. For instance, a factor can be canceled from a product in the numerator and denominator, but individual terms in a sum cannot be canceled separately. The algebraic factorization must be correct, and the equality after cancellation must be understood as applying away from the zero of the canceled factor.