Continuity of Rational Functions

Division Requires a Nonzero Denominator

The previous tutorial established continuity of polynomial functions and showed how sums and products preserve continuity. A quotient introduces one additional condition: division is defined only when the denominator is nonzero. The key local fact is that if a continuous denominator is nonzero at a point, it stays nonzero for all sufficiently nearby inputs.

The domain is part of the function’s specification. For instance, the expression \((x^2-9)/(x-3)\) is defined only when \(x\ne3\), even though its numerator can be factored and a simpler formula is available away from \(3\). A cancellation may give an equal formula on the original domain, but it does not by itself put an excluded input back into that domain.

A Nonzero Denominator Stays Away from Zero Nearby

Continuity gives more than the fact that the denominator remains nonzero. It gives a positive lower bound for its absolute value near the point. That bound is what makes division estimates possible.

Proof. Set \(c=|g(a)|\), so \(c>0\). By continuity of \(g\) at \(a\), there is a \(\delta_0>0\) such that \(|g(x)-g(a)|<c/2\) whenever \(x\in E\) and \(|x-a|<\delta_0\). The reverse triangle inequality gives \[ |g(x)|\geq |g(a)|-|g(x)-g(a)|>c-\frac{c}{2}=\frac{c}{2}. \] This proves the claim. In particular, \(g(x)\ne0\) for all such \(x\). \(\square\)

The Quotient Theorem

The local lower bound controls the denominator in a quotient difference. Continuity of the numerator and denominator then controls the two changing quantities in its numerator.

Proof. Write \(c=|g(a)|>0\). By the preceding theorem, choose \(\delta_0>0\) such that \(|g(x)|>c/2\) whenever \(x\in E\) and \(|x-a|<\delta_0\). Fix \(\varepsilon>0\). Continuity of \(f\) gives \(\delta_f>0\) such that \[ |x-a|<\delta_f\quad\Longrightarrow\quad |f(x)-f(a)|<\frac{\varepsilon c}{4} \] for \(x\in E\). If \(f(a)\ne0\), continuity of \(g\) also gives \(\delta_g>0\) such that \[ |x-a|<\delta_g\quad\Longrightarrow\quad |g(x)-g(a)|<\frac{\varepsilon c^2}{4|f(a)|}. \] If \(f(a)=0\), no additional condition involving \(\delta_g\) is needed. Choose \(\delta\) to be the minimum of \(\delta_0\), \(\delta_f\), and \(\delta_g\) when \(f(a)\ne0\); when \(f(a)=0\), choose \(\delta=\min\{\delta_0,\delta_f\}\). For \(x\in E\) with \(|x-a|<\delta\), the denominator \(g(x)\) is nonzero. The identity \[ \frac{f(x)}{g(x)}-\frac{f(a)}{g(a)} = \frac{g(a)(f(x)-f(a))-f(a)(g(x)-g(a))}{g(x)g(a)} \] and the triangle inequality imply \[ \left|\frac{f(x)}{g(x)}-\frac{f(a)}{g(a)}\right| \leq \frac{|f(x)-f(a)|}{|g(x)|} + \frac{|f(a)|\,|g(x)-g(a)|}{|g(x)|c}. \] The first term is less than \((\varepsilon c/4)/(c/2)=\varepsilon/2\). If \(f(a)\ne0\), the second term is less than \[ \frac{|f(a)|}{(c/2)c}\,\frac{\varepsilon c^2}{4|f(a)|} =\frac{\varepsilon}{2}. \] If \(f(a)=0\), the second term is zero. Thus in either case the quotient difference is less than \(\varepsilon\). This proves continuity at \(a\). \(\square\)

Continuity of Rational Functions on Their Domains

The quotient theorem applies directly to polynomials: the numerator and denominator are continuous everywhere, and the theorem applies at each point where the denominator is nonzero. The natural domain also has a useful geometric property.

Proof. Fix \(a\in D_q\), so \(q(a)\ne0\). Polynomials are continuous at every real number by the result of the previous tutorial. Apply the nonzero-denominator theorem to \(q\) at \(a\). It provides \(\delta_0>0\) such that \(q(x)\ne0\) whenever \(|x-a|<\delta_0\). Hence \((a-\delta_0,a+\delta_0)\subseteq D_q\). Since this holds for every \(a\in D_q\), the set \(D_q\) is open. At any \(a\in D_q\), both \(p\) and \(q\) are continuous at \(a\), and \(q(a)\ne0\). The quotient theorem therefore shows that \(p/q\) is continuous at \(a\). If \(D_q\) is empty, the openness assertion holds and there are no points at which to check continuity. \(\square\)

This theorem is a domain-specific continuity statement: it does not claim that a rational function is defined at zeros of its denominator. If one restricts a rational function to any subset \(E\subseteq D_q\), it remains continuous at every point of \(E\), since the continuity estimates on \(D_q\) also work when inputs are restricted to \(E\).

Worked Estimates and Domain Checks

How to Use the Quotient Rule Carefully

To apply the continuity theorem, check the point before manipulating the quotient. First verify that the point belongs to the function’s domain, which means the denominator is nonzero there. Then use polynomial continuity for the numerator and denominator and apply the quotient theorem. This order prevents a common mistake: treating a simplified expression as though it had automatically changed the original domain.

The local lower bound in the proof explains why the hypothesis \(g(a)\ne0\) matters. Near such a point, the denominator cannot become arbitrarily small, so changes in the numerator and denominator can be controlled together. At a zero of the denominator, that argument is unavailable, and the quotient may be undefined or may fail to have a continuous extension. The theorem makes no claim at such a point.