Difficulty: Intermediate | Prerequisites: Basic trigonometric identities, definition of a limit, Section 2.2–2.3 notes.
Trigonometric limits rely on one core identity: lim(x→0) sin(x)/x = 1. Nearly every trig limit in this course can be reduced to that identity through algebraic manipulation, conjugate multiplication, or rewriting other trig functions in terms of sine and cosine. Master the pattern and the algebra follows.
Fundamental trigonometric limit
lim(x→0) sin(x)/x = 1. This is the building block for every trig limit in the section. Think of it as: when the angle is tiny, sin(x) and x are nearly identical.
Indeterminate form
An expression like 0/0 that cannot be evaluated by direct substitution. In simple terms, it means the limit might still exist, but you need algebraic work to find it.
Conjugate multiplication
Multiplying numerator and denominator by a conjugate expression (e.g. √s + 6 when √s − 6 appears) to eliminate a radical or simplify an indeterminate form. Think of it as: creating a difference of squares so the troublesome term cancels.
Continuity at a point
A function f is continuous at x = a when three conditions hold: f(a) is defined, lim(x→a) f(x) exists, and lim(x→a) f(x) = f(a). In simple terms, you can draw the graph through that point without lifting your pen.
Squeeze theorem (sandwich theorem)
If g(x) ≤ f(x) ≤ h(x) near a, and lim g(x) = lim h(x) = L, then lim f(x) = L. This is the theoretical backbone behind lim sin(x)/x = 1.
Every problem in this section follows one pattern: rewrite the expression so that lim(x→0) sin(x)/x = 1 can be applied. The mechanics vary, but the goal is always the same.
When the argument of sin differs from the denominator, multiply and divide to create a matching pair.
Example: lim(x→0) sin(4x)/(3x)
Rewrite as [sin(4x)/(4x)] × (4x/3x)
The first factor → 1 by the fundamental limit
The second factor simplifies to 4/3
Answer: 4/3
When a square root appears inside sin, multiply numerator and denominator by the conjugate to create a difference of squares.
Example: lim(s→36) sin(√s − 6)/(s − 36)
Factor: s − 36 = (√s − 6)(√s + 6)
Rewrite as [sin(√s − 6)/(√s − 6)] × 1/(√s + 6)
First factor → 1; second factor → 1/(6 + 6) = 1/12
Answer: 1/12
tan(x) = sin(x)/cos(x), so any limit involving tan can be decomposed.
Example: lim(x→0) tan(x)/(10x)
Rewrite as [sin(x)/(x)] × [1/(10 cos(x))]
First factor → 1; second factor → 1/(10 × 1) = 1/10
Answer: 1/10
When sin has a polynomial argument, match the argument inside and outside sin, then simplify the remaining rational expression.
Example: lim(x→0) sin(x² + 5x)/(3x)
Rewrite as [sin(x² + 5x)/(x² + 5x)] × [(x² + 5x)/(3x)]
First factor → 1; second factor simplifies to (x + 5)/3
As x → 0, this gives (0 + 5)/3 = 5/3
Answer: 5/3
When the limit point is not zero, factor the argument of sin to create a form that approaches zero.
Example: lim(u→2) sin(u² − 4)/(u − 2)
Factor: u² − 4 = (u − 2)(u + 2)
Rewrite as [sin((u − 2)(u + 2))/((u − 2)(u + 2))] × (u + 2)
As u → 2, the first factor → 1 and (u + 2) → 4
Answer: 4
Problem 4 asks: for the piecewise function f(x) = sin(−5x)/x when x < 0, and f(x) = x − a when x ≥ 0, find a so that f is continuous at x = 0.
Left-hand limit: lim(x→0⁻) sin(−5x)/x = −5 × [sin(−5x)/(−5x)] = −5 × 1 = −5
Right-hand limit: lim(x→0⁺) (x − a) = 0 − a = −a
For continuity, set −a = −5, so a = 5
\lim_{x \to 0} \frac{\sin x}{x} = 1\lim_{x \to 0} \frac{1 - \cos x}{x} = 0\tan x = \frac{\sin x}{\cos x}Continuity condition: f is continuous at x = a when
\lim_{x \to a} f(x) = f(a)The identity lim sin(x)/x = 1 underpins the small-angle approximation used throughout physics and engineering: for small angles (in radians), sin(θ) ≈ θ. This is why pendulum period formulas, lens equations, and structural load calculations simplify so neatly at small deflections.
Students often think lim sin(x)/x = 1 works for degrees. It does not. The identity requires x in radians.
Students sometimes cancel sin from sin(4x)/(4x) as if sin were a factor. sin is a function, not a multiplier.
When the limit point is not zero (e.g. u → 2), students try to apply sin(u)/u = 1 directly. The identity only works when the argument of sin approaches zero, so you must factor first.
Students forget to check all three continuity conditions and assume that matching left and right limits is enough. The function value f(a) must also equal the limit.
⚠️ The sin(x)/x limit is almost guaranteed to appear on Exam 2, often disguised with a different argument (sin(3x)/x, sin(x²)/x, etc.).
⚠️ Continuity problems that combine piecewise functions with trig limits are a staple. Expect to solve for an unknown constant.
⚠️ Problems may mix radicals with trig (the conjugate technique), so practise both in combination.
⚠️ If you finish early, check whether your final answer is a clean fraction. Most answers in this section are simple ratios.
True or false: lim(x→0) sin(x)/x = 0.
False. The limit equals 1.
Fill in the blank: lim(x→0) sin(7x)/(7x) = ___.
True or false: To evaluate lim sin(√s − 6)/(s − 36), you can substitute s = 36 directly.
False. Direct substitution gives 0/0, an indeterminate form.
Fill in the blank: tan(x) = sin(x) / ___.
cos(x).
True or false: For a piecewise function to be continuous at a point, only the left and right limits need to match.
False. The function value at that point must also equal the common limit.
Q: Evaluate lim(x→0) sin(4x)/(3x).
A: Rewrite as [sin(4x)/(4x)] × (4/3). The first factor → 1, so the answer is 4/3.
Q: Evaluate lim(s→36) sin(√s − 6)/(s − 36).
A: Factor s − 36 = (√s − 6)(√s + 6). The limit becomes [sin(√s − 6)/(√s − 6)] × 1/(√s + 6) → 1 × 1/12 = 1/12.
Q: Evaluate lim(x→0) tan(x)/(10x).
A: Rewrite tan(x) as sin(x)/cos(x). The limit becomes [sin(x)/x] × [1/(10 cos(x))] → 1 × 1/10 = 1/10.
Q: Evaluate lim(x→0) sin(x² + 5x)/(3x).
A: Rewrite as [sin(x² + 5x)/(x² + 5x)] × [(x² + 5x)/(3x)]. The first factor → 1, and (x² + 5x)/(3x) = (x + 5)/3 → 5/3.
Q: Find the value of a that makes f(x) = sin(−5x)/x for x < 0 and f(x) = x − a for x ≥ 0 continuous at x = 0.
A: The left-hand limit is −5. The right-hand limit is −a. Setting −a = −5 gives a = 5.
This material connects directly to the derivative of sin(x), which you will prove using lim sin(h)/h = 1 in the next section on differentiation rules. It also connects to L'Hôpital's rule later in the course, which provides an alternative method for indeterminate forms, though you are expected to use algebraic techniques for now.
trigonometric limits, sin x over x, sinc function, squeeze theorem, sandwich theorem, small-angle approximation, indeterminate form 0/0, conjugate multiplication, piecewise continuity, MTH 132, Calculus I, Michigan State, Section 2.4, limit of trig functions, fundamental trig limit