Why Composite Sine Ratios Matter
Many trigonometric limits do not contain a sine of \(x\) alone. The angle may instead be an expression such as \(3x-6\), \(x^2-4\), or \(\sqrt{x+3}-2\). The important question is whether that entire angle approaches zero, and how quickly it does so compared with the denominator.
The standard sine limit provides a reliable way to handle these expressions. Once the angle is treated as a single quantity, the problem becomes a comparison between that quantity and the denominator. This also explains why a factor such as \(3\) or \(4\) can remain in the answer: the angle and denominator may approach zero at different rates.
The Composite Sine Limit
The standard sine limit is \(\lim_{u\to 0}\sin(u)/u=1\), with \(u\) measured in radians. A composite expression uses a function \(g(x)\) in place of the variable \(u\). If \(g(x)\to 0\) as \(x\to a\), the input to the sine function approaches the value needed for the standard limit.
The final sentence matters because a quotient \(\sin(g(x))/g(x)\) is not literally defined when \(g(x)=0\). The limit concerns values arbitrarily close to \(a\), not necessarily the value at \(a\). Assigning the value \(1\) at zero is valid for reasoning with the limit, since \(\sin(0)=0\) and the extended ratio is continuous there.
For a quotient with another denominator, a useful form is to separate the sine ratio from the comparison of the two small quantities. Suppose \(h(x)\ne 0\) for \(x\) sufficiently close to \(a\), except possibly at \(a\), and suppose \(g(x)/h(x)\to L\). Then the composite sine quotient approaches \(L\).
Why the Denominator Comparison Works
The proof is an exact factorization followed by the product limit law. When \(g(x)\ne 0\), write the quotient as a product. If \(g(x)=0\), use the extended value \(1\) for the sine ratio; the product identity still holds because both \(\sin(g(x))\) and \(g(x)\) equal zero.
Factor the expression into the standard sine ratio and the comparison of the angle with the denominator: \(\sin(g(x))/h(x)=[\sin(g(x))/g(x)][g(x)/h(x)]\), using the extended sine ratio when \(g(x)=0\).
Since \(g(x)\to 0\), the first factor tends to \(1\). By assumption, the second factor tends to \(L\).
The product tends to \(1\cdot L=L\), so \(\sin(g(x))/h(x)\to L\).
This derivation shows what the method requires. The angle must tend to zero, and its ratio to the denominator must have a finite limit. Merely observing that both the numerator and denominator approach zero is not enough to determine the quotient.
Worked Examples
Worked Example: A Linear Composite Angle
Evaluate \(\lim_{x\to 2}\sin(3x-6)/(x-2)\).
Let \(g(x)=3x-6\) and \(h(x)=x-2\). As \(x\to 2\), \(g(x)\to 3(2)-6=0\), and \(h(x)\to 0\). For \(x\ne 2\), factor the quotient:
The first factor tends to \(1\) by the composite sine limit, and the second factor is \(3\). Therefore the limit is \(1\cdot 3=3\).
Check: The comparison \(g(x)/h(x)=(3x-6)/(x-2)=3\) for \(x\ne 2\), so the theorem gives the same result, \(3\).
Worked Example: A Quadratic Composite Angle
Evaluate \(\lim_{x\to 2}\sin(x^2-4)/(x-2)\).
Set \(g(x)=x^2-4\). Then \(g(x)\to 2^2-4=0\). Factor the difference of squares to compare the angle with the denominator:
As \(x\to 2\), the sine ratio tends to \(1\), while \(x+2\to 4\). Thus the limit is \(1\cdot 4=4\).
Check: For \(x\ne 2\), \(g(x)/(x-2)=x+2\), whose limit is \(4\). This agrees with the result.
Worked Example: A Radical Composite Angle
Evaluate \(\lim_{x\to 1}\sin(\sqrt{x+3}-2)/(x-1)\).
Let \(g(x)=\sqrt{x+3}-2\). As \(x\to 1\), \(g(x)\to\sqrt{4}-2=0\). Direct division by \(g(x)\) is useful, but first rationalize \(g(x)\) to compare it with \(x-1\):
for \(x\ne 1\). Therefore \(g(x)/(x-1)\to 1/(2+2)=1/4\). Now factor the original expression:
The first factor tends to \(1\), and the second tends to \(1/4\), so the limit is \(1\cdot 1/4=1/4\).
Check: The computed comparison ratio \(g(x)/(x-1)=1/(\sqrt{x+3}+2)\) has limit \(1/4\); multiplying by the sine-ratio limit \(1\) confirms the answer.
How to Choose the Comparison
In practice, the main work is often simplifying \(g(x)/h(x)\), not handling the sine factor. Start by identifying the full angle \(g(x)\), then verify that it approaches zero. Next, compare it with the denominator \(h(x)\), using ordinary algebra such as factoring or rationalization when needed.
- If \(g(x)/h(x)\to L\), the composite sine quotient tends to \(L\).
- If \(g(x)/h(x)\to 0\), the sine quotient tends to \(0\).
- If \(g(x)/h(x)\) does not have a finite limit, this result alone does not determine a finite limit for the sine quotient.
The last point is a limit on what the theorem tells us, not a claim that the original expression can never have a limit. A separate argument may be required if the comparison ratio fails to converge.
Common Mistake: Losing the Scale Factor
A frequent error is to conclude that a quotient approaches \(1\) whenever its sine angle and denominator both approach zero. The standard sine limit only says that \(\sin(g(x))/g(x)\to 1\), not that \(\sin(g(x))/h(x)\to 1\) for every vanishing denominator.
Worked Example: Correcting a Scale-Factor Error
Consider \(\lim_{x\to 0}\sin(5x)/x\). A mistaken calculation might say, “\(5x\to 0\), so the limit is \(1\).” That omits the comparison between \(5x\) and \(x\). The correct factorization is:
As \(x\to 0\), the first factor tends to \(1\), and the second is \(5\). The limit is \(5\), not \(1\).
Check: The angle-to-denominator ratio is \(5x/x=5\) for \(x\ne 0\). The theorem therefore predicts \(5\), confirming the corrected calculation.
Radians Are Essential
The standard sine limit used here assumes that the angle is measured in radians. If angles are measured in degrees, the numerical value of \(\sin(u)/u\) near zero is not \(1\), because the degree measure and radian measure use different scales. In AP Calculus, trigonometric limit formulas are used with radian measure.
Also distinguish the sine ratio from a related expression. The conclusion that \(\sin(g(x))/g(x)\to 1\) does not by itself evaluate a quotient with an unrelated denominator. Always display the factor \(g(x)/h(x)\); it makes the scale visible and helps prevent a missing constant.
Summary
To evaluate a composite sine ratio, first check that its angle \(g(x)\) tends to zero. Then split the quotient into \(\sin(g(x))/g(x)\) and \(g(x)/h(x)\). The first factor tends to \(1\), while the second records how the angle compares with the denominator. If that comparison tends to a finite value \(L\), the original quotient tends to \(L\).
Check Your Understanding
For each limit, identify the composite angle and its ratio to the denominator before stating the result. Use radians.
- Evaluate \(\lim_{x\to 3}\sin(2x-6)/(x-3)\).
- Evaluate \(\lim_{x\to 1}\sin(x^2-1)/(x-1)\).
- Evaluate \(\lim_{x\to 0}\sin(4x+x^2)/x\).
- In the theorem for \(\sin(g(x))/h(x)\), why must \(g(x)/h(x)\) also be considered?
- What is the error in concluding that \(\lim_{x\to 0}\sin(7x)/(2x)=1\) just because \(7x\to 0\)?
