Q.A rational function is a function of the form $f(x) = \dfrac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomial functions of $x$ and $q(x) \neq 0$. Prove that every such rational function is continuous (i.e. continuous at every point of its domain, the set of all real $x$ for which $q(x) \neq 0$).

Concept: Continuity At A Point — A function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For rational functions, we use the fact that polynomials are continuous everywhere.

Step 1: Polynomials $p(x)$ and $q(x)$ are continuous for all real $x$ (standard result: $\lim_{x \to a} p(x) = p(a)$, same for $q$).

Step 2: For any $a$ in the domain (i.e. $q(a) \neq 0$), the quotient rule for limits applies:

Step 3: Since the limit equals the function value at every $a$ where $q(a) \neq 0$, $f$ is continuous at every point of its domain.

The key idea is that continuity is preserved under the usual algebraic operations — addition, subtraction, multiplication, and division (provided the denominator is non‑zero). Since polynomials are continuous everywhere, a rational function inherits continuity wherever its denominator does not vanish.

Let’s walk through the reasoning step by step.

1. Polynomials are continuous everywhere.

   A polynomial $p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$ is built from the constant function and the identity function $x$ using only addition and multiplication. Both $c$ (constant) and $x$ are continuous at every real number. Repeated application of the limit laws — the sum and product of continuous functions are continuous — shows that any polynomial is continuous for all $x \in \mathbb{R}$.

2. The quotient of two continuous functions is continuous where the denominator is non‑zero.

   This is a standard theorem: if $g$ and $h$ are both continuous at $x = a$, and $h(a) \neq 0$, then the function $\frac{g}{h}$ is also continuous at $x = a$. The proof uses the limit law for quotients:

provided the denominator limit is non‑zero. This is exactly the definition of continuity at $a$.

3. Apply this to a rational function.

   Let $f(x) = \frac{p(x)}{q(x)}$, where $p$ and $q$ are polynomials. For any real number $a$ such that $q(a) \neq 0$:

   • $p$ is continuous at $a$ (by step 1).
   • $q$ is continuous at $a$ (by step 1).
   • Since $q(a) \neq 0$, the quotient rule applies, so $f$ is continuous at $a$.

4. What about points where $q(a) = 0$?

   Those points are not in the domain of $f$. Continuity is only defined at points where the function itself is defined. So the statement “every rational function is continuous” means: it is continuous at every point of its domain. There is no requirement to consider points outside the domain.

Method: Proving a Quotient of Two Function Families Is Continuous on Its Domain

This method applies whenever a function is built as one continuous function divided by another (rational functions being the standard example), and you must establish continuity everywhere the division is actually valid.

Steps

Step 1: Establish continuity of the numerator and denominator separately.

Show (or cite as already proven) that both the numerator $p(x)$ and the denominator $q(x)$ are continuous at every real number — for polynomials this follows from the algebra of continuous functions built from constants and the identity function.

Step 2: Invoke the quotient rule for continuity.

This is the key theorem that turns "numerator and denominator are each continuous" into "the ratio is continuous," but only where the denominator doesn't vanish.

Step 3: Restrict the conclusion to the actual domain.

Continuity can only be asked about at points where the function is defined. So identify exactly the set where $q(x) \neq 0$ — that is the domain — and state the result as "continuous at every point of the domain," never as "continuous everywhere" without qualification.

Step 4: Do not treat zeros of the denominator as discontinuities.

A point where $q(a)=0$ is simply outside the domain; the function isn't discontinuous there in the technical sense, because discontinuity requires the point to be a domain point where continuity fails, not a point where the function doesn't exist at all.

Common Mistakes

Mistake 1: Stating the conclusion as "continuous everywhere" instead of "continuous on its domain."

Why it's wrong: a rational function is undefined wherever the denominator is zero, so it can never be continuous "everywhere" in the literal sense (e.g. $f(x)=1/x$ has no value at $x=0$) — the correct, precise claim is continuity at every point of the domain. Correct approach: always attach the domain qualifier when stating the conclusion for a rational function.

Mistake 2: Calling the denominator's zero a "point of discontinuity."

Why it's wrong: a point excluded from the domain entirely (division by zero) is not a discontinuity in the formal sense, just a gap in the domain — discontinuity is only meaningful at a point that fails one of the three continuity conditions while still plausibly belonging to the function. Correct approach: describe such points as "not in the domain," reserving "discontinuous" for domain points where the limit fails to match the value.

Mistake 3: Forgetting to justify that the numerator and denominator are continuous before applying the quotient rule.

Why it's wrong: the quotient rule is only valid once both pieces are already known to be continuous — skipping straight to "so the quotient is continuous" hides a real logical gap. Correct approach: explicitly state that $p(x)$ and $q(x)$ are polynomials, hence continuous by the algebra of continuous functions, before invoking the quotient theorem.