Why the Denominator Matters
A rational function is a quotient of polynomials. Its numerator and denominator are each continuous for every real input, but the quotient is defined only where its denominator is nonzero. That restriction gives a direct way to identify where continuity is guaranteed—and which inputs require separate attention.
The key distinction is between a function’s domain and a formula that might simplify algebraically. Cancelling a common factor can make a formula easier to evaluate at some inputs, but it does not change the domain of the original function. If the original denominator was zero at an input, the original function remains undefined there.
Domain and Continuity for a Rational Function
A polynomial \(P(x)\) is continuous for every real number \(x\). For polynomials \(P\) and \(Q\), the rational function \(R\) is defined by
at inputs where \(Q(x)\ne 0\). Thus its natural domain is the set of real inputs for which the denominator is not zero.
This result follows from the quotient rule for continuity: if \(P\) and \(Q\) are continuous at \(a\) and \(Q(a)\ne 0\), then \(\frac{P}{Q}\) is continuous at \(a\). Both polynomial conditions hold for every real \(a\); the nonzero-denominator condition is the restriction that remains.
Why the Quotient Rule Guarantees Continuity
To be continuous at \(a\), a function must be defined at \(a\), have a limit as \(x\) approaches \(a\), and have that limit equal its value at \(a\). Suppose \(P\) and \(Q\) are polynomials and \(Q(a)\ne 0\). Polynomial continuity gives the limits of the numerator and denominator directly. The quotient limit law applies because the denominator’s limit is nonzero.
The last equality also confirms that \(R(a)\) is defined: \(Q(a)\ne 0\). Therefore the limit exists, the function value exists, and they are equal. By the three-condition definition of continuity, \(R\) is continuous at \(a\). Since this argument works for every \(a\) in the natural domain, the rational function is continuous throughout that domain.
Worked Examples
Worked Example: Finding the Continuity Set from the Denominator
Find the natural domain and the set of points where \(f(x)=\frac{x^2+2x+5}{x^2-1}\) is continuous.
The denominator must be nonzero. Factor it and find its zeros:
At \(x=1\), the denominator is \(1^2-1=0\). At \(x=-1\), it is \((-1)^2-1=0\). These two inputs are excluded. At every other real input, the denominator is nonzero, so the quotient continuity theorem applies.
Check: At \(x=0\), the denominator is \(0^2-1=-1\ne 0\), so the theorem guarantees continuity there. At \(x=1\) and \(x=-1\), the denominator is zero, so neither input belongs to the domain. The result excludes exactly the two denominator zeros.
Worked Example: Cancellation Does Not Change the Original Domain
For \(g(x)=\frac{x^2-9}{x-3}\), find the natural domain and state where the original function is continuous. Then examine the excluded input.
The original denominator is zero when \(x-3=0\), so \(x=3\) is excluded. For every \(x\ne 3\), factor the numerator and simplify:
Therefore the original function has domain \(\mathbb{R}\setminus\{3\}\) and, by the quotient continuity theorem, is continuous at every input in that domain. The simplified formula agrees with the original only where the original quotient is defined.
At the excluded input, the original expression would require division by zero:
so \(g(3)\) is undefined. However, for \(x\ne 3\), the simplified expression is \(x+3\), giving
Check: Substituting \(x=3\) into the simplified formula gives \(3+3=6\), which matches the limit. It does not give the value of the original function: that function remains undefined at \(3\). The continuity theorem establishes continuity for \(x\ne 3\), not at the excluded input.
Worked Example: Distinguishing Two Excluded Inputs
For \(h(x)=\frac{(x-2)(x+1)}{(x-2)(x-4)}\), find its natural domain and continuity set, then examine the behavior at each excluded input.
The original denominator is zero at \(x=2\) and \(x=4\). Thus the domain is \(\mathbb{R}\setminus\{2,4\}\), and the quotient continuity theorem guarantees continuity at every point in this domain.
For \(x\ne 2\) and \(x\ne 4\), the common factor can be cancelled:
At the first excluded input, the simplified expression has a finite limit:
But \(h(2)\) is undefined, since the original denominator is \((2-2)(2-4)=0\cdot(-2)=0\). At the other excluded input, the simplified expression has a denominator approaching zero while its numerator approaches \(5\). The one-sided behavior is
Indeed, near \(4\), \(x+1\) is positive; \(x-4\) is negative on the left and positive on the right. The quotient therefore has opposite signs on the two sides, with unbounded magnitude.
Check: The original denominator equals zero at both \(2\) and \(4\), so both are excluded, and the continuity set is exactly \(\mathbb{R}\setminus\{2,4\}\). The finite limit at \(2\) and the unbounded one-sided behavior at \(4\) are different kinds of behavior, but neither makes the original function continuous at an input where it is undefined.
What Denominator Zeros Do—and Do Not—Tell You
Solving \(Q(x)=0\) gives every input where the original quotient is undefined. It does not, by itself, determine what happens near those inputs. At an excluded input, the numerator might also be zero, allowing a factor to cancel and leaving a finite limit. Or the numerator might remain nonzero while the denominator approaches zero, producing unbounded behavior. These possibilities require separate analysis, such as factoring and evaluating a limit or examining one-sided signs.
The crucial point for this topic is that every denominator zero is excluded from the original domain, even if later algebra reveals a finite limit there. The quotient continuity theorem applies only where the original denominator is nonzero. Analysis at an excluded input can describe the limit, but it cannot retroactively make the original function defined at that input.
Write down \(Q(x)\) as it appears in the original function, before cancelling factors.
Solve \(Q(x)=0\). Every real solution is excluded from the natural domain.
Because the numerator and denominator are polynomials, the quotient is continuous wherever \(Q(x)\ne 0\).
If a limit or local behavior is requested, examine each excluded input separately. Do not include it in the original domain merely because a factor cancels.
A Common Mistake
A frequent error is to cancel a factor and then treat the simplified expression as if it were the original function everywhere. For example, this calculation is wrong:
The cancellation divides by \(x-3\), so it is valid only when \(x\ne 3\). The incorrect last claim includes \(x=3\), even though the original denominator is zero there. The correction is
At \(x=3\), the original function is undefined; the simplified expression can help find a limit but cannot change that fact. Always determine the domain from the original denominator before simplifying.
Why This Rule Is Useful
For rational functions, continuity can often be established without calculating a limit separately at every ordinary input. Once the real zeros of the denominator are known, the quotient theorem guarantees continuity at every other real input. This makes the domain the organizing feature of the analysis: it identifies both where the continuity rule applies and where it does not.
When an exercise asks where a rational function is continuous, give the set of inputs explicitly. Saying only “where the denominator is nonzero” describes a rule; listing the excluded inputs shows that you have applied it. If a denominator factors, use those factors to make the exclusions clear, and keep the original formula in view when considering cancellation.
Check Your Understanding
For each question, use the original denominator to identify the domain before considering any cancellation.
- Find the natural domain and continuity set of \(r(x)=\frac{x+4}{x^2-16}\).
- For \(s(x)=\frac{x^2-4}{x-2}\), what is the domain of the original function, and what is its limit as \(x\to 2\)?
- State the condition on \(Q(a)\) needed to conclude that \(P(x)/Q(x)\) is continuous at \(a\).
- For \(t(x)=\frac{x+1}{x^2+1}\), explain why the quotient continuity theorem applies at every real input.
- If a common factor cancels and the simplified expression has a value at an excluded input, does that value belong to the original function? Explain.
