Check the Conditions Before Using a Trigonometric Limit
Standard trigonometric limits are powerful, but they apply only when their conditions are met. In particular, the familiar sine limit requires an angle approaching zero, measured in radians. A quotient that contains sine or cosine is not automatically an instance of that limit.
Before calculating, ask three questions: What does the trigonometric argument approach? Does the denominator approach zero, and if so, at what rate compared with the numerator? Are the angles measured in radians? These checks help distinguish a finite limit from an expression that grows without bound or has no two-sided limit.
The Standard Limits and Their Requirements
The two basic limits used in this topic are stated for real angles measured in radians:
In both limits, the variable \(u\) must approach zero. The first limit says that \(\sin u\) and \(u\) have a ratio approaching 1 near zero. The second says that \(1-\cos u\) behaves like \(\frac12u^2\) near zero. Neither limit says that \(\sin u/u\) is always 1, or that \((1-\cos u)/u^2\) is always \(\frac12\).
If a function \(g(x)\) approaches zero, one may use the sine limit on \(\sin(g(x))/g(x)\). The matching denominator \(g(x)\) matters: if the expression has a different denominator, it must be related to \(g(x)\) before the standard limit can help. Continuity can also be useful when the trigonometric argument approaches a nonzero value. In that case, the trig function approaches its value at that point; the zero-angle standard limits do not apply.
When the Angle Does Not Approach Zero
Suppose \(x\) approaches \(a\), and the angle \(g(x)\) approaches a nonzero value \(c\). Then continuity gives \(\sin(g(x))\to\sin c\) and \(\cos(g(x))\to\cos c\). It is incorrect to replace a quotient involving these expressions by a standard zero-angle limit without first establishing that the angle approaches zero.
Worked Example: A Nonzero Sine Numerator Over a Vanishing Denominator
Evaluate, or determine that there is no finite two-sided limit:
Step 1: Check the angle. As \(x\to0\), the angle \(2+x\to2\), not zero. Thus \(\sin(2+x)\to\sin2\), rather than behaving like \(2+x\). Since \(2\) radians lies between \(0\) and \(\pi\), \(\sin2>0\).
Step 2: Check the denominator and signs. The denominator \(x\to0\), while the numerator approaches the positive number \(\sin2\). For \(x>0\) sufficiently close to zero, the quotient is positive and its magnitude grows without bound. For \(x<0\) sufficiently close to zero, the quotient is negative and its magnitude grows without bound.
Conclusion and check: The one-sided behaviors have opposite signs, so the two-sided limit does not exist as a finite value. The sign check is consistent with the original expression: its numerator stays positive near zero, while its denominator changes sign there.
This example illustrates why the numerator’s limit matters. A denominator approaching zero does not by itself make an expression an indeterminate form suitable for a standard trig limit. Here the numerator approaches a nonzero value, so there is no cancellation of the kind seen in \(\sin u/u\).
When Both Numerator and Denominator Approach Zero
If both numerator and denominator approach zero, the standard limits may help, but the rates of approach still matter. For example, \(\sin x\) is comparable to \(x\) near zero in radians. If the denominator is \(x^2\), there is an extra factor of \(1/x\), so the quotient does not approach a finite number.
Worked Example: The Denominator Vanishes Faster Than the Sine
Determine whether the following limit is finite:
For \(x\ne0\), factor the expression as
The standard sine limit gives \(\sin x/x\to1\). For \(x>0\), the first factor is positive and approaches 1, while \(1/x\to+\infty\); the product grows without bound. For \(x<0\) sufficiently close to zero, \(\sin x/x\) is still positive, while \(1/x\to-\infty\).
Conclusion and check: There is no finite two-sided limit. The factorization verifies the diagnosis: the sine-to-angle ratio stays near 1, leaving the factor \(1/x\) to determine the unbounded behavior and its sign.
The sine limit is used correctly here, but it does not produce the answer by itself. Rewriting the expression exposes the extra factor and shows why a finite limit is impossible. This is an important distinction: a standard limit can be part of a calculation without being the entire calculation.
Angles Measured in Degrees
The standard limits above use radians. If an angle is given in degrees, convert it before applying them. A degree measure \(x^\circ\) corresponds to the radian measure \(\frac{\pi x}{180}\). Forgetting this conversion changes the value of the limit.
Worked Example: A Sine Limit with a Degree-Measured Angle
Let \(x\) denote a numerical degree measure. Evaluate:
Step 1: Convert the angle. In radians, \(x^\circ=\frac{\pi x}{180}\). Therefore
Step 2: Match the sine limit’s denominator. Multiply and divide by \(\frac{\pi x}{180}\):
Step 3: Take the limit. As \(x\to0\), the radian angle \(\frac{\pi x}{180}\to0\). The standard sine limit applies to the second factor:
Conclusion and check: The limit is \(\frac{\pi}{180}\), not 1. The conversion factor remains after the sine-to-angle ratio tends to 1, confirming the result.
A Reliable Rescaling Check
A constant multiple in the angle does not prevent use of a standard limit. If the angle still approaches zero, rewrite the expression so it contains the standard ratio. For the cosine defect, this means matching the denominator to the square of the angle.
Worked Example: A Rescaled Cosine Defect
Evaluate:
Here \(3x\to0\), and the cosine-defect limit uses \((3x)^2\) in the denominator. Since \((3x)^2=9x^2\), rewrite:
As \(x\to0\), the second factor tends to \(\frac12\). Thus
Conclusion and check: The limit is \(\frac92\). The factor 9 is required because the standard limit has the squared angle \((3x)^2\), not \(x^2\), in its denominator.
Why the Conditions Matter
The sine limit comes from the behavior of sine near zero: in radians, \(\sin u/u\) approaches 1 as \(u\to0\). If the angle does not approach zero, that local relationship cannot be substituted into the expression. If the denominator is not the angle—or the square of the angle for the cosine defect—an additional factor may remain. If the angle is measured in degrees, the conversion factor must also be retained.
These checks are especially useful when a limit has a familiar-looking trig function but an unfamiliar surrounding expression. A sound approach is to identify the angle, find its limiting value, inspect the denominator, and only then decide whether a standard limit can be used.
If it is not zero, do not apply a zero-angle standard limit to that angle.
If the denominator approaches zero but the numerator approaches a nonzero value, do not treat the quotient as a standard sine or cosine limit.
When the angle approaches zero, rewrite using that angle in the denominator, or its square for the cosine defect, and keep any leftover factors.
Convert degree measures to radians before using the standard limits.
A Common Mistake
A common error is to assume that \(\sin u/u\to1\) whenever a quotient contains sine and a denominator that approaches zero. For example, one might look at \(\sin(2+x)/x\) and claim that its limit is 1 because it resembles \(\sin u/u\).
This does not match the standard limit: the sine’s angle \(2+x\) approaches 2, not zero, and the denominator \(x\) is not \(2+x\). In fact, the numerator approaches \(\sin2>0\), while the denominator approaches zero and changes sign. The corrected conclusion is that the right-hand quotient grows without bound, the left-hand quotient decreases without bound, and the finite two-sided limit does not exist.
The corresponding degree mistake is to claim \(\lim_{x\to0}\sin(x^\circ)/x=1\). The correct conversion gives \(\frac{\pi}{180}\), because the radian angle is \(\frac{\pi x}{180}\). In both cases, the cure is the same: check the exact argument and units, then rewrite the expression rather than relying on its appearance.
Check Your Understanding
For each question, identify whether a standard trigonometric limit applies and explain the key condition you checked.
- As \(x\to0\), what does the angle \(4-x\) approach? Can the standard sine limit be applied directly to \(\sin(4-x)\)?
- Does \(\lim_{x\to0}\frac{\sin x}{x^2}\) have a finite two-sided value? Which factorization helps diagnose it?
- When \(x\) is a numerical degree measure, what radian angle corresponds to \(x^\circ\)?
- In \(\frac{1-\cos(5x)}{x^2}\), what factor is needed to match the squared-angle denominator?
- Why is the statement \(\sin u/u\to1\) not enough, by itself, to evaluate every quotient containing sine?
