Difficulty: Intermediate | Prerequisites: Basic limit laws, unit circle values, algebra of fractions.
Trigonometric limits sit at the bridge between the limit concept (Sections 2.1 to 2.3) and derivatives, which you will meet in full in Chapter 3. The centrepiece is the special limit sin(x)/x → 1 as x → 0, which underpins the derivative formulas for sin, cos, tan and the rest. If you are comfortable with limit laws and basic trig identities, you are ready for this material. If not, revisit the unit circle and the algebra of rational expressions first.
Many limits involving trig functions look indeterminate (0/0) at first glance. The strategy is almost always to rewrite the expression so that the core identity lim(x→0) sin(x)/x = 1 can be applied, then simplify. Mastering this one identity, plus a few algebraic tricks, covers the vast majority of exam problems on trig limits.
Fundamental trig limit (sin x / x)
lim(x→0) sin(x)/x = 1. This is the single most important identity for evaluating trig limits. In simple terms, when the angle is tiny, sin of that angle is almost exactly equal to the angle itself (in radians).
Indeterminate form 0/0
An expression that evaluates to 0/0 when you plug in the limiting value directly. Think of it as a signal: you cannot just substitute, you need algebraic work first.
Conjugate multiplication
Multiplying numerator and denominator by the conjugate of a radical expression to clear the radical. In simple terms, this is the trick of using (a + b)(a − b) = a² − b² to simplify expressions with square roots.
Continuity at a point
A function f is continuous at x = c when lim(x→c) f(x) = f(c). Think of it as: you can draw the graph through that point without lifting your pen.
Squeeze Theorem (context)
If g(x) ≤ f(x) ≤ h(x) near c, and lim g(x) = lim h(x) = L, then lim f(x) = L. This is the theorem behind the proof of sin(x)/x → 1, though in practice you use the result rather than re-proving it on exams.
Everything flows from two results:
lim(x→0) sin(x)/x = 1
lim(x→0) (1 − cos(x))/x = 0
The second follows from the first via conjugate multiplication. If you know these two cold, you can handle nearly every trig limit on a Calculus I exam.
When you see a trig limit that gives 0/0, your job is to rewrite it so that sin(something)/something appears, where the "something" approaches 0. The steps:
Factor constants out of the limit.
If the argument inside sin differs from the denominator, multiply and divide to match them.
Apply lim sin(u)/u = 1 as u → 0.
Evaluate whatever remains by direct substitution.
Example (a): lim(x→0) sin(4x) / (3x)
Rewrite as (1/3) · [sin(4x) / x]. Then multiply and divide by 4: (4/3) · [sin(4x) / (4x)]. As x → 0, the bracketed part → 1.
Answer: 4/3.
Example (b): lim(s→36) sin(√s − 6) / (s − 36)
Factor the denominator: s − 36 = (√s − 6)(√s + 6). Cancel the common factor:
sin(√s − 6) / [(√s − 6)(√s + 6)].
Let u = √s − 6, so u → 0 as s → 36. This becomes [sin(u)/u] · [1/(√s + 6)].
As s → 36: sin(u)/u → 1 and 1/(√36 + 6) = 1/12.
Answer: 1/12.
Example (c): lim(x→0) tan(x) / (10x)
Write tan(x) = sin(x)/cos(x). Then tan(x)/(10x) = (1/10) · [sin(x)/x] · [1/cos(x)].
As x → 0: sin(x)/x → 1, and 1/cos(0) = 1.
Answer: 1/10.
Example (d): lim(x→0) sin(x² + 5x) / (3x)
Factor inside the sine: x² + 5x = x(x + 5). So sin(x(x + 5))/(3x).
Rewrite as (1/3) · [sin(x(x + 5)) / (x(x + 5))] · (x + 5).
As x → 0: the sin/argument ratio → 1, and (x + 5) → 5.
Answer: 5/3.
Example (e): lim(u→2) sin(u² − 4) / (u − 2)
Factor: u² − 4 = (u − 2)(u + 2). So sin((u − 2)(u + 2)) / (u − 2).
Rewrite as [sin((u − 2)(u + 2)) / ((u − 2)(u + 2))] · (u + 2).
Let t = (u − 2)(u + 2); as u → 2, t → 0, so sin(t)/t → 1, and (u + 2) → 4.
Answer: 4.
A classic exam question gives you a piecewise function and asks you to find a constant that makes it continuous at a particular point. The method:
Compute the left-hand limit (x approaching from below).
Compute the right-hand limit (x approaching from above).
Set them equal to each other and to the function value at that point.
Solve for the unknown.
f(x) = sin(−5x)/x for x < 0, and f(x) = x − a for x ≥ 0.
Find the value of a that makes f continuous at x = 0.
Left-hand limit (x → 0⁻): sin(−5x)/x = −5 · [sin(−5x)/(−5x)]. As x → 0, sin(−5x)/(−5x) → 1, so the left-hand limit is −5.
Right-hand limit and function value at x = 0: f(0) = 0 − a = −a.
Set them equal: −a = −5, so a = 5.
The key insight: the trig limit identity does the heavy lifting on the left side, and then continuity just becomes an algebra problem.
Students often think lim sin(x)/x = 1 works for any value of x. It does not. It only holds as x → 0. If x → π, you just substitute directly.
Students frequently forget to match the argument of sine to the denominator. Writing sin(4x)/x = 1 is wrong. You must have sin(4x)/(4x) = 1, then account for the extra factor of 4 separately.
Students sometimes apply the identity to cosine: lim cos(x)/x as x → 0 is not 1. Cos(0) = 1, so cos(x)/x → 1/0, which diverges. The companion identity is (1 − cos x)/x → 0.
When a denominator factors (as in s − 36 = (√s − 6)(√s + 6)), students sometimes try L'Hopital's Rule instead of factoring. At this stage of MTH 132, factoring and the sin(u)/u identity are the expected tools.
⚠️ The identity lim sin(x)/x = 1 is tested in almost every Calculus I midterm, often more than once.
⚠️ Expect at least one problem where the argument inside sin does not match the denominator and you must introduce a correction factor.
⚠️ Piecewise continuity problems ("find a so that f is continuous") are a favourite exam question because they combine trig limits with the definition of continuity.
⚠️ Be ready for variants using tan(x) = sin(x)/cos(x). Rewriting tan in terms of sin and cos is a standard first step.
True or False: lim(x→0) sin(7x)/(7x) = 1.
Fill in the blank: lim(x→0) sin(3x)/(5x) = ___.
True or False: lim(x→0) cos(x)/x = 1.
Fill in the blank: To evaluate lim(x→0) tan(x)/(2x), the first step is to rewrite tan(x) as ___.
True or False: If f is defined piecewise and lim(x→c⁻) f(x) = lim(x→c⁺) f(x) = f(c), then f is continuous at c.
Answers: 1. True. 2. 3/5. 3. False (it diverges). 4. sin(x)/cos(x). 5. True.
Q: Evaluate lim(x→0) sin(6x) / (2x).
A: Rewrite as (6/2) · sin(6x)/(6x) = 3 · 1 = 3.
Q: Evaluate lim(x→0) sin(x²) / x.
A: Rewrite as x · [sin(x²)/x²]. As x → 0, sin(x²)/x² → 1, and the leading x → 0. The limit is 0.
Q: Evaluate lim(t→0) tan(3t) / (7t).
A: Write tan(3t) = sin(3t)/cos(3t). Then (1/7) · [sin(3t)/t] · [1/cos(3t)] = (1/7)(3)(1) = 3/7.
Q: Find the value of b that makes g(x) continuous at x = 0, where g(x) = sin(2x)/x for x < 0 and g(x) = 3x + b for x ≥ 0.
A: Left-hand limit: sin(2x)/x = 2 · sin(2x)/(2x) → 2. At x = 0: g(0) = b. Set b = 2.
Q: Evaluate lim(x→9) sin(√x − 3) / (x − 9).
A: Factor x − 9 = (√x − 3)(√x + 3). Cancel, giving [sin(√x − 3)/(√x − 3)] · 1/(√x + 3). As x → 9 the first factor → 1, the second → 1/6. Answer: 1/6.
This material feeds directly into the derivative of sin(x). In Chapter 3, you will see that the limit definition of the derivative of sin(x) at any point requires exactly the identity lim(h→0) sin(h)/h = 1. If you are solid on trig limits now, the derivative proofs will feel almost automatic.
The piecewise continuity problems connect to Section 2.3 (continuity) and reappear in later chapters when you study differentiability, since a function must be continuous before it can be differentiable.
Related Terms / Search Tags
trig limits, trigonometric limits, sin x over x, sine over x limit, lim sinx/x, fundamental trig limit, special trig limits, limit involving sine, limit involving tangent, piecewise continuity, continuity at a point, 0/0 indeterminate form, MTH 132, Calculus I, Michigan State University, Section 2.4