Why a Nonzero Denominator Matters
A quotient can inherit continuity from its numerator and denominator, but only where the denominator is nonzero. This qualification is essential: division by zero is undefined, so the quotient is not even a function at an input where its denominator vanishes.
The useful conclusion is local. If both functions are continuous at a point and the denominator has a nonzero value there, then their quotient is continuous at that point. If the denominator is zero, this result gives no conclusion. The quotient may be undefined there, even if algebraic simplification makes a related expression look harmless.
Domains and the Quotient Theorem
Suppose \(f:A\to\mathbb{R}\) and \(g:B\to\mathbb{R}\). The quotient \(f/g\) can be defined only at inputs belonging to both original domains where \(g\) is nonzero. Its domain is therefore
Continuity at a point is defined for a function at an input in its domain. In epsilon-delta form, a function \(h\) is continuous at \(a\in D_h\) if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that for every \(x\in D_h\),
The condition \(g(a)\ne 0\) does two jobs. It ensures that the quotient has a value at \(a\), and it allows us to keep the denominator away from zero for inputs sufficiently close to \(a\). We prove both the local bound and the continuity estimate below.
Why the Quotient Theorem Works
Write \(q(x)=f(x)/g(x)\) wherever \(g(x)\ne0\). Since \(g(a)\ne0\), continuity of \(g\) at \(a\) lets us make \(g(x)\) close enough to \(g(a)\) that \(g(x)\) cannot be zero. In particular, we can require
The reverse triangle inequality then gives
Now compare the quotient at \(x\) with its value at \(a\). Combining the fractions and using the bound on \(|g(x)|\), we obtain
Both differences on the final line can be made small because \(f\) and \(g\) are continuous at \(a\). Here are explicit choices. Given \(\varepsilon>0\), choose a positive \(\delta\) small enough to ensure all three conditions:
Such a \(\delta\) exists: continuity supplies a suitable positive distance for each condition, and we take the minimum of those distances. The first condition gives the denominator bound. The other two, substituted into the estimate, give
This is exactly the epsilon-delta condition for \(f/g\) to be continuous at \(a\). It also confirms that \(g(x)\ne0\) for all sufficiently close \(x\), so the quotient is defined throughout that puncture-free neighborhood within the common domain. The proof depends on the nonzero value of \(g(a)\); without it, the denominator bound cannot be obtained.
Applying the Theorem at Allowed Inputs
Worked Example: A Quotient of Two Polynomials
Consider \(q(x)=(x^2+1)/(x+2)\). Determine whether it is continuous at \(x=1\), and find its value there.
The numerator \(f(x)=x^2+1\) and denominator \(g(x)=x+2\) are polynomials, so both are continuous at \(1\). Evaluate the denominator:
The quotient theorem applies. Substitution gives
Thus \(q\) is continuous at \(1\), with value \(2/3\). Check the conditions at the point: the numerator equals \(2\), the denominator equals \(3\), and their quotient is \(2/3\), as stated. The full domain excludes \(x=-2\), where the denominator is zero.
Worked Example: A Denominator Root Away from the Input
Let \(r(x)=(x+2)/(x^2-1)\). Is \(r\) continuous at \(x=0\)?
Both the numerator and denominator are polynomials, hence continuous at \(0\). The denominator at that input is
Therefore \(r\) is continuous at \(0\), and direct substitution gives
The result is checked by evaluating the original expression at \(0\): its numerator is \(2\), its denominator is \(-1\), and their quotient is \(-2\). The expression is undefined at \(x=1\) and \(x=-1\), but those are not the input being tested. Their presence in the overall domain restriction does not prevent continuity at \(0\).
Forbidden Inputs: Where the Theorem Does Not Apply
A denominator zero is a forbidden input for the quotient as written. It is important to be precise about what follows: the quotient theorem cannot establish continuity at such an input, because its hypothesis \(g(a)\ne0\) fails. If the quotient is undefined there, it cannot be continuous there as a function with that original domain.
Sometimes algebraic cancellation produces a simpler formula that has a value at the forbidden input. Cancellation can describe the quotient at other inputs, but it does not change the original domain or assign a missing value. The distinction is especially visible in the next example.
Worked Example: Cancellation Does Not Define the Missing Value
Consider \(s(x)=(x^2-1)/(x-1)\). Is \(s\) continuous at \(x=1\)?
At \(x=1\), the denominator is \(1-1=0\). Thus \(s(1)\) is undefined, and \(1\) is excluded from the domain. The quotient theorem does not apply there, since its denominator condition fails.
For \(x\ne1\), factor the numerator and cancel the common factor:
This equality is restricted to \(x\ne1\). It shows that values of \(s(x)\) near \(1\) agree with \(x+1\), but it does not make \(s(1)\) equal to \(2\). The original function remains undefined at \(1\), so it is not continuous there. One could define a new function by assigning the value \(2\) at \(1\); that would be an extension, not the original quotient.
Check the domain distinction directly: substituting \(1\) into the original expression gives \(0/0\), which is undefined, whereas substituting into the simplified expression gives \(2\). The simplification was valid only for \(x\ne1\), so these calculations do not conflict.
A Common Mistake
A frequent error is to cancel a factor and then claim that the original quotient is continuous at the input where that factor vanished.
Another error is to check only that the denominator is nonzero somewhere nearby. The theorem requires that it be nonzero at the point in question, as well as requiring continuity of both functions there. For example, even though \(x-3\) is continuous, the expression \(1/(x-3)\) is not defined at \(x=3\), so the theorem cannot be used to claim continuity at \(3\).
Using the Result Carefully
For a quotient, the continuity test at a particular input is short, but each condition matters. Identify the input and verify that it belongs to the domains of both component functions. Check continuity of the numerator and denominator there. Then evaluate the denominator at that exact input. If it is nonzero, the theorem applies; if it is zero, the quotient is excluded there and this result does not apply.
Confirm that the numerator and denominator are both defined at \(a\).
Verify that both component functions are continuous at \(a\).
If \(g(a)\ne0\), the quotient theorem proves continuity at \(a\). If \(g(a)=0\), the quotient is not defined there as written.
The quotient theorem is useful because it lets us establish continuity without returning to the epsilon-delta definition each time. Its domain condition also prevents a misleading shortcut: a formula that simplifies away a factor may still retain the original expression’s excluded input.
Check Your Understanding
For each question, state whether the quotient theorem applies at the specified input and explain why.
- If \(f\) and \(g\) are continuous at \(a\) and \(g(a)=5\), what can you conclude about \(f/g\) at \(a\)?
- For \(u(x)=(x+1)/(x^2+4)\), is the quotient theorem applicable at \(x=0\)? Identify the denominator value.
- For \(v(x)=(x^2-9)/(x-3)\), what is the domain restriction, and does cancellation assign a value at \(x=3\)?
- Why does the quotient proof need continuity of the denominator as well as the condition \(g(a)\ne0\)?
- If a denominator vanishes at the input, what can and cannot be concluded from the quotient theorem?
