Reciprocal Bounds and the Quotient Law Proof

Why a Nonzero Limit Matters

The quotient law says that if \(f(x)\) approaches \(L\) and \(g(x)\) approaches \(M\), with \(M\ne 0\), then \(f(x)/g(x)\) approaches \(L/M\). The condition \(M\ne 0\) is essential to the proof: it lets us show that \(g(x)\) cannot be too close to zero when \(x\) is sufficiently close to the limit point.

That nearby lower bound is the key step. Once the denominator is bounded away from zero, taking its reciprocal is controlled, and the quotient law follows by applying the product law to \(f(x)\) and \(1/g(x)\).

Limits and a Nearby Lower Bound

For a function \(f\) defined near \(a\), the statement \(\lim_{x\to a}f(x)=L\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that

for all \(x\) in the domain of \(f\). The limit concerns values near \(a\); \(f\) need not be defined at \(a\).

Suppose \(L\ne0\). In the definition of the limit, choose the allowed error to be \(|L|/2\). Then sufficiently close to \(a\), the value \(f(x)\) differs from \(L\) by less than half the magnitude of \(L\). The triangle inequality gives a positive lower bound for \(|f(x)|\).

Proof of the Bound

Because \(f(x)\to L\), the limit definition with error \(|L|/2>0\) supplies a \(\delta_0>0\) for which

The triangle inequality, applied in the form \(|L|\le |L-f(x)|+|f(x)|\), now gives

This proves the bound. The final quantity is positive, so \(|f(x)|>0\) and therefore \(f(x)\ne0\). The function may have a zero at \(a\) itself, if it is defined there; the result only guarantees nonzero values in a sufficiently small punctured neighborhood.

Worked Example: A Reciprocal Near a Positive Limit

The Reciprocal Law

The nearby lower bound allows us to control the difference between the reciprocal and its proposed limit. Notice the exact identity

The numerator becomes small because \(f(x)\to L\). The lower bound on \(|f(x)|\) prevents the denominator from becoming arbitrarily small. Together, these facts prove the reciprocal law.

To prove the limit, let \(\varepsilon>0\) be given. Because \(f(x)\to L\), there is a positive distance within which \(|f(x)-L|\) is less than any prescribed positive tolerance. Use the tolerance

Choose \(\delta>0\) so that \(0<|x-a|<\delta\) implies \(|f(x)-L|<\eta\). The first part of the minimum ensures \(|f(x)|>|L|/2\), so the reciprocal is defined there. Then

Thus the definition of the limit is satisfied, proving the reciprocal law. The argument explains why a nonzero \(L\) is required: if \(L=0\), the lower bound \(|f(x)|>|L|/2\) would give no positive separation from zero, and \(1/L\) would not be defined.

Worked Example: A Reciprocal Near a Negative Limit

The Quotient Law

The product law for limits states that if \(f(x)\to L\) and \(h(x)\to H\), then \(f(x)h(x)\to LH\). When \(g(x)\to M\ne0\), the reciprocal law gives \(1/g(x)\to1/M\). In a sufficiently small punctured neighborhood, \(g(x)\ne0\), so

Applying the product law to the right-hand side proves the quotient law.

The nonzero condition ensures more than a meaningful proposed answer \(L/M\). By the bound theorem, \(g(x)\) is nonzero nearby, so the quotient itself is defined there. It also ensures that the reciprocal law applies.

Worked Example: Applying the Quotient Law

A Common Mistake: Ignoring the Nonzero Condition

A tempting but invalid step is to apply the quotient law when the denominator limit is zero. For example, \(x/x\) is defined for \(x\ne0\), but as \(x\to0\), both its numerator and denominator approach zero. The quotient law does not apply: its denominator-limit condition fails. In this particular example, \(x/x=1\) whenever \(x\ne0\), so its limit is \(1\). That conclusion follows from simplifying the function on its punctured domain, not from the quotient law.

The reciprocal version shows why the condition cannot simply be dropped. Since \(x\to0\), one might try to conclude that \(1/x\to1/0\), but \(1/0\) is not a real number, and \(1/x\) does not have a finite real limit at \(0\). Indeed, suppose it had a finite limit \(C\). Using the limit definition with error \(1\), there would be a \(\delta>0\) such that \(0<x<\delta\) implies \(|1/x-C|<1\). Choose \(x\) positive with \(x<\min(\delta,1/(|C|+2))\). Then \(1/x>|C|+2\), so

contradicting \(|1/x-C|<1\). Thus there is no finite real limit. The correct habit is to check the denominator’s limiting value before invoking the quotient law.