Where a Quotient Is Defined
For a product, values of both factors are enough to define the result throughout their common domain. A quotient has an additional restriction: its denominator must not vanish. That restriction is central to continuity. At a single point, a nonzero denominator gives pointwise continuity when the numerator and denominator are continuous. Across an entire domain, a denominator that stays a positive distance from zero gives stronger control.
The quotient continuity theorem from “Continuity of Rational Functions” states that if \(f\) and \(g\) are continuous at \(a\in E\) and \(g(a)\ne0\), then \(f/g\) is continuous at \(a\), as a function defined wherever its denominator is nonzero. Thus continuity of the factors does not make the quotient defined at points where \(g=0\). The theorem applies only at points where the quotient exists.
The algebraic identity below is useful for stronger, domain-wide conclusions. For \(x,y\in D\), subtract the quotient values by first changing the numerator while keeping one denominator fixed:
Indeed, expanding the right side gives \(f(x)/g(x)-f(y)/g(x)+f(y)/g(x)-f(y)/g(y)\); the middle terms cancel. Since \[ \left|\frac{1}{g(x)}-\frac{1}{g(y)}\right| =\frac{|g(x)-g(y)|}{|g(x)|\,|g(y)|}, \] the triangle inequality yields
This estimate separates the changes in the numerator and denominator. It also shows why the denominator must be controlled: small denominators can magnify either change.
A Positive Lower Bound Gives Uniform Continuity
For uniform continuity, a pointwise condition \(g(x)\ne0\) at each point is not by itself the convenient global control needed in the estimate. The useful condition is that there be one number \(m>0\) such that \(|g(x)|\geq m\) everywhere. Suppose also that the numerator is bounded, so changes in the denominator cannot be magnified by arbitrarily large numerator values.
Proof. If \(E\) is empty, uniform continuity holds vacuously. Otherwise, the lower bound on \(|g|\) ensures \(g(x)\ne0\) for every \(x\in E\), so the quotient is defined throughout \(E\). Fix \(\varepsilon>0\), and set
Here \(A>0\), \(B\geq0\), and \(\eta>0\). By uniform continuity of \(f\), there is a \(\delta_f>0\) such that \(x,y\in E\) and \(|x-y|<\delta_f\) imply \(|f(x)-f(y)|<\eta\). Uniform continuity of \(g\) gives \(\delta_g>0\) such that the same input condition with \(\delta_g\) implies \(|g(x)-g(y)|<\eta\). Take \(\delta=\min\{\delta_f,\delta_g\}\). For \(x,y\in E\) with \(|x-y|<\delta\), the quotient estimate and the hypotheses give
The strict inequality in the third line follows because both changes are less than \(\eta\); it remains valid if \(B=0\), since \(A>0\). The last inequality uses \(A+B<A+B+1\). Thus this \(\delta\) verifies uniform continuity of \(f/g\). \(\square\)
Worked Example: A Quotient That Is Lipschitz on a Closed Interval
Let \(E=[-1,1]\), \(f(x)=x^2+1\), and \(g(x)=x+3\). For \(x\in E\), \(1\leq f(x)\leq2\), so \(f\) is bounded by \(M=2\). Also \(g(x)\geq2\), so \(|g(x)|\geq m=2\). For \(x,y\in E\),
Thus both functions are Lipschitz, and hence uniformly continuous. The theorem shows that \(f/g\) is uniformly continuous on \(E\). More precisely, the Lipschitz quotient estimate proved below gives the constant \[ \frac{2}{2}+\frac{2\cdot1}{2^2}=\frac32. \] For example, the quotient takes the values \(f(1)/g(1)=2/4=1/2\) and \(f(-1)/g(-1)=2/2=1\). Their difference is \(1/2\), which is at most \((3/2)|1-(-1)|=3\), as the bound requires.
A Quantitative Lipschitz Estimate
The same calculation gives a numerical rate of control. Recall that \(h:E\to\mathbb{R}\) is Lipschitz with constant \(L\geq0\) if \(|h(x)-h(y)|\leq L|x-y|\) for all \(x,y\in E\). A positive lower bound for the denominator controls the two denominator factors in the quotient estimate.
Proof. For any \(x,y\in E\), apply the quotient estimate. The lower bound on \(|g|\), the bound on \(|f|\), and the two Lipschitz inequalities give
This is the Lipschitz inequality with the stated constant. It does not require the denominator to be bounded above. \(\square\)
A useful special case concerns reciprocals. Taking \(f(x)=1\) gives a bounded Lipschitz numerator with \(M=1\) and \(L_f=0\). Therefore, if \(g\) is Lipschitz and \(|g(x)|\geq m>0\) on \(E\), then \(1/g\) is Lipschitz with constant \(L_g/m^2\). In particular, it is uniformly continuous.
When the Denominator Approaches Zero
The pointwise quotient theorem requires the denominator to be nonzero at the point under consideration. This gives continuity there, but it does not promise uniform continuity across a domain where the denominator can become arbitrarily close to zero. The following example verifies the failure directly, including the domain of every input.
Worked Example: The Reciprocal on an Open Interval
Define \(q(x)=1/x\) on \(E=(0,1)\). The reciprocal is continuous at every point of \(E\), by the pointwise quotient theorem (or the reciprocal continuity result established earlier). But \(q\) is not uniformly continuous on \(E\). For each integer \(n\geq2\), take
Both inputs belong to \(E\): \(0<1/(n+1)<1/n\leq1/2<1\). Their distance is
The quotient values are \(q(x_n)=n\) and \(q(y_n)=n+1\), so \[ |q(x_n)-q(y_n)|=|n-(n+1)|=1 \] for every \(n\geq2\). To see the failure from the definition, choose \(\varepsilon=1/2\). Given any \(\delta>0\), choose an integer \(n\geq2\) large enough that \(n(n+1)>1/\delta\). Then \(|x_n-y_n|=1/[n(n+1)]<\delta\), but \(|q(x_n)-q(y_n)|=1>1/2\). No single \(\delta\) works for this \(\varepsilon\). The denominator \(x\) is nonzero throughout \(E\), yet it has no positive lower bound there.
A positive lower bound is a sufficient condition for the uniform-continuity theorem, not a necessary condition for every quotient to be uniformly continuous. For instance, the quotient of the constant function \(1\) by itself is constant and uniformly continuous on any domain. The lower-bound hypothesis is valuable because it guarantees control for general numerator and denominator functions, rather than relying on special cancellations.
Cancellation and the Domain
Worked Example: A Continuous Extension Does Not Fill a Missing Point Automatically
On \(E=\mathbb{R}\setminus\{1\}\), define \[ r(x)=\frac{x^2-1}{x-1}. \] The denominator is nonzero for every \(x\in E\). Factoring the numerator gives \(x^2-1=(x-1)(x+1)\); hence, for each \(x\in E\),
Thus \(r\) is continuous at every point of its domain, since it agrees there with the continuous function \(x\mapsto x+1\). The formula \(x+1\) also suggests assigning the value \(2\) at \(x=1\), which would produce a continuous extension to all of \(\mathbb{R}\). But the original quotient is not defined at \(1\): its numerator and denominator are both zero there, and cancellation does not change that fact. The extension is a new function with a larger domain.
These examples distinguish three questions. First, is the quotient defined at the point? Second, if it is defined, are the numerator and denominator continuous there with a nonzero denominator? The pointwise quotient theorem answers the continuity question affirmatively. Third, for a uniform or Lipschitz conclusion, do we have global bounds that prevent small denominators or large numerator values from magnifying changes?
Check Your Understanding
Use the definitions, estimates, and hypotheses in this tutorial to answer the following questions.
- On what set is \(f/g\) defined when \(f,g:E\to\mathbb{R}\)?
- In the uniform-continuity theorem, which hypotheses control the two denominator factors, and which hypothesis controls the numerator value multiplying the denominator change?
- Suppose \(|f|\leq4\), \(|g|\geq2\), and the Lipschitz constants of \(f\) and \(g\) are \(3\) and \(5\), respectively. What Lipschitz constant does the quotient theorem guarantee?
- For the sequence example on \((0,1)\), why must the index be restricted to \(n\geq2\), and what are the input distances and output differences?
- Why does cancelling \(x-1\) in \((x^2-1)/(x-1)\) not make the original quotient defined at \(x=1\)?