Learning Objectives
Use the fundamental sine limit with radian inputs.
Rewrite scaled angles so a known limit becomes visible.
Use identities to evaluate limits involving 1 − cos x.
Prerequisites: the Squeeze Theorem, radian measure, and the identities sin²x + cos²x = 1 and sin(−x) = −sin x. This lesson does not use derivatives or L'Hôpital's rule.
1. The Fundamental Sine Limit
With x measured in radians, lim(x → 0) (sin x)/x = 1. The quotient is undefined at zero, so direct substitution produces 0/0. The limit describes how closely sin x and x agree for small nonzero angles.
For 0 < x < π/2, a unit-circle area comparison gives sin x < x < tan x. Dividing and rearranging yields cos x < (sin x)/x < 1. Since cos x approaches 1, the Squeeze Theorem gives the right-hand limit 1. The quotient is even, because sin(−x)/(−x) = sin x/x, so the left-hand limit is also 1.
Common Mistakes
Radians are essential. If an input d is measured in degrees, sin(d degrees)/d approaches π/180, not 1.
The statement is about a quotient near zero. It does not assign a value to 0/0.
2. Match the Angle and Denominator
Worked Example: A Scaled Angle
Evaluate lim(x → 0) sin(5x)/(2x). Rewrite the quotient as (5/2)[sin(5x)/(5x)].
As x approaches zero, 5x also approaches zero. The bracketed quotient tends to 1.
The limit is 5/2. Rewriting with 5x in the denominator preserves the known sine-limit pattern; the outside factor keeps the expression equivalent.
For constants a and b with b ≠ 0, this approach gives lim(x → 0) sin(ax)/(bx) = a/b. When a = 0, the numerator is identically zero and the same conclusion follows directly.
Worked Example: A Ratio of Sines
Evaluate lim(x → 0) sin(3x)/sin(7x). For sufficiently small nonzero x, rewrite it as [sin(3x)/(3x)] × [7x/sin(7x)] × (3/7).
The first factor tends to 1, and the second is the reciprocal of a quotient tending to 1.
The limit is 3/7. Check that the limiting denominator is nonzero before taking a reciprocal.
3. Use a Conjugate with Cosine
Worked Example: A Second-Order Limit
Evaluate lim(x → 0) (1 − cos x)/x². Multiply by (1 + cos x)/(1 + cos x).
The numerator becomes 1 − cos²x = sin²x. Thus the expression equals [(sin x)/x]²/(1 + cos x).
The numerator tends to 1 and the denominator tends to 2. The limit is 1/2.
This also gives lim(x → 0) (1 − cos x)/x = 0: write the quotient as x times (1 − cos x)/x². Separately, tan x/x = (sin x/x)/cos x approaches 1, since cos x approaches 1.
4. Practice
1. Find lim(x → 0) sin(4x)/x.
2. Find lim(x → 0) tan(2x)/(3x).
3. Find lim(x → 0) (1 − cos(3x))/x².
4. Find lim(x → 0) x/sin(6x).
Worked Solutions
1. Write 4[sin(4x)/(4x)]. The limit is 4.
2. Write (2/3)[tan(2x)/(2x)]. The bracket tends to 1, so the result is 2/3.
3. Rewrite as 9[1 − cos(3x)]/(3x)². The known cosine limit gives 9/2. The factor is 9 rather than 3 because the denominator is squared.
4. Write (1/6)[6x/sin(6x)]. The reciprocal sine quotient tends to 1, so the limit is 1/6.
5. Summary
- Use lim(sin x/x) = 1 only with radian inputs.
- Match the expression inside the sine with the denominator, compensating outside.
- For 1 − cos x, a conjugate converts the numerator into sin²x.
- Check scale factors carefully when a denominator is squared.