Products of Vanishing Trigonometric Factors

Why Products of Vanishing Factors Need Care

When \(x\) approaches zero, both \(\sin(3x)\) and \(\sin(5x)\) approach zero. It is tempting to conclude that their product, even after division by \(x^2\), must also approach zero. But the denominator approaches zero too, so direct substitution produces the indeterminate form \(0/0\). It does not determine the limit.

The useful fact is that each sine factor vanishes at a rate proportional to \(x\). More precisely, for a fixed real constant \(a\), the scaled sine limit describes the ratio of \(\sin(ax)\) to \(ax\). Applying that result to both factors reveals how their constants combine.

The Scaled Sine Limit and the Product Formula

Angles in the standard sine limit must be measured in radians. The scaled sine limit states that for a fixed nonzero real number \(a\),

Since \(a\) is constant and nonzero, this also gives

This form is useful when two sine factors are divided by \(x^2\). For fixed real constants \(a\) and \(b\), the product limit is

When both constants are nonzero, the key factorization is

The equality follows by multiplying the constants and denominators: \(ab(ax)^{-1}(bx)^{-1}\) is not needed; directly, \(ab\cdot \frac{\sin(ax)}{ax}\cdot\frac{\sin(bx)}{bx}\) has denominator \(abx^2\), which cancels the leading \(ab\), leaving the original expression. Equivalently, each ratio tends to \(1\), while the fixed factor \(ab\) remains. The product law for limits therefore gives the result.

Why the Factorization Works

The proof relies on the scaled sine limit and the product law for limits. The product law says that if two functions have finite limits \(L\) and \(M\), then their product has limit \(LM\). Here the scaled ratios both approach \(1\), so their product approaches \(1\); multiplying by \(ab\) gives \(ab\). The factorization is valid for \(x\ne 0\), which is sufficient because a limit at zero concerns values near zero, not necessarily the value at zero.

This proves the formula for all fixed real \(a\) and \(b\). The special zero-constant case matters because a ratio such as \(\sin(ax)/(ax)\) is not defined when \(a=0\); it should not be used in that case.

Worked Examples

Zero Coefficients and Domain Details

If \(a=0\), then \(\sin(ax)=\sin(0)=0\) for every real \(x\). For \(x\ne0\), the quotient \(\frac{\sin(0)\sin(bx)}{x^2}\) is therefore zero, and its limit is zero. The same reasoning applies if \(b=0\). This agrees with the formula \(ab\), which is zero in either case.

At \(x=0\), the displayed quotient has denominator zero and is not defined. That does not prevent its limit from existing: limits describe behavior at nearby inputs, and the quotient is defined for every \(x\ne0\) sufficiently close to zero. Do not substitute \(x=0\) into the original quotient to decide the limit.

Common Mistake: Treating \(0/0\) as Zero

A tempting but incorrect calculation for \(\frac{\sin(3x)\sin(5x)}{x^2}\) is to substitute \(x=0\), see two zero sine factors, and conclude that the limit is zero. The substitution actually produces \(0/0\), which is indeterminate: it gives no conclusion about the limit.

Another error is to use \(\lim_{x\to0}\sin(ax)=0\) for each factor and multiply those limits, without accounting for division by \(x^2\). The denominator also tends to zero. The correction is to distribute one factor of \(x\) to each sine factor:

When \(a\) and \(b\) are nonzero, these factors tend to \(a\) and \(b\), respectively, because \(\frac{\sin(ax)}{x}=a\frac{\sin(ax)}{ax}\). Their product tends to \(ab\). This version makes clear why the numerator approaching zero does not force the entire quotient to approach zero.

How to Recognize and Use the Pattern

For a quotient with two sine factors and denominator \(x^2\), identify the fixed multipliers inside the sine functions. If the numerator is \(\sin(ax)\sin(bx)\), record \(a\) and \(b\), including any negative signs. Then apply the product formula. If there is an additional constant outside the quotient, multiply the result by that constant.

• The arguments must have the form \(ax\) and \(bx\), with \(a\) and \(b\) fixed as \(x\) approaches zero.
• The denominator \(x^2\) supplies one factor of \(x\) for each sine factor.
• The standard scaled sine limit assumes angles are measured in radians.
• A zero coefficient makes its sine factor identically zero and can be handled directly.

The result also helps explain the behavior of these expressions: near zero, each sine factor is approximately its argument, so the numerator behaves like \((ax)(bx)=abx^2\). Dividing by \(x^2\) leaves \(ab\). The exact limit proof, however, comes from the scaled sine ratios rather than from treating this approximation as an equality.