Why Quotients Need an Extra Condition
A quotient depends on both its numerator and its denominator. If the denominator approaches a nonzero number, its values stay away from zero sufficiently close to the target. That makes it possible to combine the numerator and denominator limits into a limit for the quotient.
If the denominator instead approaches zero, the quotient law gives no conclusion. The quotient might have a limit, or it might not; the behavior depends on how both functions approach their limits. The nonzero condition is therefore part of the law, not a minor technical detail.
The Quotient Limit Law
Limits here are taken as \(x\) approaches a target \(a\). Assume \(f\) and \(g\) are defined for \(x\) sufficiently close to \(a\), with \(x\ne a\), and that both limits below exist as finite real numbers.
The conclusion concerns values near \(a\), not necessarily the value at \(a\). A function or quotient may be undefined at the target and still have a limit there, provided it is defined at nearby inputs as required.
Why a Nonzero Limit Keeps the Denominator Away from Zero
Suppose \(g(x)\to M\) as \(x\to a\), with \(M\ne 0\). By the definition of a limit, for every positive tolerance \(\varepsilon\), there is a positive distance \(\delta\) such that \(0<|x-a|<\delta\) implies \(|g(x)-M|<\varepsilon\). Choose the particular tolerance \(\varepsilon=|M|/2\). The triangle inequality then gives the crucial bound.
Thus \(g(x)\ne 0\) throughout that punctured neighborhood. The function \(g\) might still equal zero at \(x=a\), if it is defined there; the bound applies to nearby \(x\ne a\).
Proof of the Quotient Limit Law
We now justify not only that the quotient is defined nearby, but also that it approaches \(L/M\). The proof uses the denominator bound above and the assumed limits of \(f\) and \(g\).
Since \(g(x)\to M\ne 0\), choose a neighborhood where \(|g(x)|>|M|/2\). Therefore \(g(x)\ne 0\), and \(f(x)/g(x)\) is defined there.
For \(x\) in that neighborhood, rearrange the difference using a common denominator.
The resulting expression is controlled by \(|f(x)-L|\) and \(|g(x)-M|\), both of which become small near \(a\).
In detail, whenever \(g(x)\ne 0\) and \(M\ne 0\),
Using the triangle inequality and \(|g(x)|>|M|/2\), we obtain
Let \(\varepsilon>0\) be any desired tolerance for the quotient. Since \(f(x)\to L\), we can make the first term less than \(\varepsilon/2\) by requiring \(|f(x)-L|<\varepsilon|M|/4\). If \(L\ne 0\), since \(g(x)\to M\), we can make the second term less than \(\varepsilon/2\) by requiring \(|g(x)-M|<\varepsilon|M|^2/(4|L|)\). If \(L=0\), the second term is zero and needs no restriction. Choose a single neighborhood small enough to satisfy these requirements and the earlier bound \(|g(x)-M|<|M|/2\). Then the displayed difference is less than \(\varepsilon\). This proves that the quotient approaches \(L/M\).
Using the Law with Known Function Limits
The quotient law can be combined with the sum and product limit laws. First find the limits of the numerator and denominator using those laws; then check that the denominator limit is nonzero before dividing.
Worked Example: A Quotient of Two Functions
Suppose, as \(x\to a\), that \(f(x)\to 6\) and \(g(x)\to -3\). Find the limit of \(\frac{f(x)}{g(x)}\).
The numerator limit is \(6\), and the denominator limit is \(-3\). Since \(-3\ne 0\), the quotient limit law applies:
The nonzero condition can also be seen directly. Because \(g(x)\to -3\), sufficiently close to \(a\) we have \(|g(x)+3|<1\). It follows that \(|g(x)|>2\), so \(g(x)\ne0\) there. As an arithmetic check, the proposed quotient limit \(-2\), multiplied by the denominator limit \(-3\), gives the numerator limit \(6\).
Worked Example: A Quotient with a Sum in the Denominator
Suppose \(u(x)\to 2\) and \(v(x)\to 5\) as \(x\to a\). Find the limit of \(\frac{3u(x)-v(x)}{u(x)+v(x)}\).
Apply the sum and product limit laws to the numerator and denominator separately:
The denominator limit is \(7\), which is nonzero. The quotient law therefore gives
To verify that the denominator stays nonzero nearby, the limit \(u(x)+v(x)\to7\) ensures that \(|u(x)+v(x)-7|<1\) sufficiently close to \(a\). Thus \(u(x)+v(x)>6\), so the denominator cannot be zero. Finally, multiplying the proposed quotient limit \(1/7\) by the denominator limit \(7\) returns the numerator limit \(1\).
Worked Example: A Denominator That Is Zero at the Target
Define \(g(x)=2\) for \(x\ne 0\), and \(g(0)=0\). Define \(f(x)=6\) for \(x\ne 0\). Find \(\lim_{x\to0} f(x)/g(x)\).
For every \(x\ne0\), \(f(x)=6\) and \(g(x)=2\). Consequently, as \(x\to0\),
The denominator limit \(2\) is nonzero, so the quotient limit law applies. For nearby \(x\ne0\), the quotient is \(6/2=3\), and hence
Although \(g(0)=0\), that does not prevent the limit from existing: the quotient is defined for every nearby \(x\ne0\). Checking directly, \(f(x)/g(x)=3\) for all such inputs, which confirms the limit.
When the Denominator Limit Is Zero
If \(\lim_{x\to a}g(x)=0\), the quotient limit law does not apply. This does not prove that the quotient has no limit; it means the law cannot determine the answer. The numerator’s behavior and the denominator’s behavior must then be examined more closely.
Worked Example: Why the Zero Condition Matters
For \(x\ne0\), let \(g(x)=x\). Compare the quotients \(f_1(x)/g(x)\) and \(f_2(x)/g(x)\), where \(f_1(x)=x\) and \(f_2(x)=2x\). As \(x\to0\), all three functions \(f_1\), \(f_2\), and \(g\) approach \(0\).
The denominator limit is \(0\), so the quotient limit law cannot be used to divide the function limits. Instead, inspect the quotients at \(x\ne0\):
Thus the first quotient approaches \(1\) and the second approaches \(2\), even though both numerators and both denominators approach \(0\). The different outcomes show why replacing a quotient by a ratio of limits is not justified when the denominator limit is zero. Directly checking the simplified expressions confirms each result.
A Common Mistake
A frequent error is to write “the numerator limit divided by the denominator limit” without first checking whether the denominator limit is zero. For example, from \(f(x)\to0\) and \(g(x)\to0\), it is incorrect to conclude
The right-hand calculation is invalid: \(0/0\) is not a defined number, and the quotient limit law requires a nonzero denominator limit. The correction is to state that the law does not apply, then analyze the quotient itself using an appropriate method. The example above shows that quotients with both component limits equal to zero can have different limits.
What to Check Before Dividing Limits
Before applying the quotient law, identify the numerator limit and denominator limit separately. Then ask whether the denominator limit is nonzero. If it is, the law applies and also guarantees the quotient is defined close to the target, excluding the target if necessary. If it is zero, do not divide the limits; the quotient requires further analysis.
Check Your Understanding
For each question, state whether the quotient limit law applies and explain the relevant denominator condition.
- If \(f(x)\to -4\) and \(g(x)\to 2\), what is \(\lim_{x\to a}f(x)/g(x)\)?
- If \(p(x)\to 3\) and \(q(x)\to -1\), find the limit of \(\bigl(p(x)+2q(x)\bigr)/q(x)\).
- Why does \(g(x)\to M\ne0\) imply that \(g(x)\ne0\) sufficiently close to the target?
- Can the quotient have a limit if its denominator is zero at the target but its denominator limit is nonzero? Explain.
- If the numerator and denominator both approach zero, what does the quotient limit law allow you to conclude?
