Difficulty: Introductory calculus | Prerequisites: Limit laws, basic trig identities, unit circle, squeeze theorem concept
Trigonometric limits are the bridge between the definition of the derivative and the trig derivative formulas you just learnt. The central result, lim(x→0) sin(x)/x = 1, is what makes the derivative of sin(x) equal cos(x). Every problem in this section is a variation on that one fact, dressed up with algebra. If you can reduce a limit to the sin(u)/u form, you can solve it.
The key limit is lim(x→0) sin(x)/x = 1. From it you also get lim(x→0) tan(x)/x = 1 and lim(x→0) (1 − cos(x))/x = 0. Every trig limit problem on this assignment can be solved by rewriting the expression so that one of these standard forms appears.
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. Think of it as: if a function is trapped between two others that both go to the same value, it has no choice but to go there too. This is the theorem used to prove lim(x→0) sin(x)/x = 1.
The fundamental trig limit: lim(x→0) sin(x)/x = 1
The single most important limit in this section. It says that for very small angles (in radians), sin(x) is approximately equal to x. In simple terms, the sine function and the identity function are nearly identical near the origin.
The complementary trig limit: lim(x→0) (1 − cos(x))/x = 0
Cosine starts at 1 and curves away from it very slowly (quadratically), so the numerator shrinks faster than x does. This limit shows up less often but is needed for certain problems.
lim(x→0) tan(x)/x = 1
Follows directly from sin(x)/x = 1 and cos(0) = 1, since tan(x)/x = (sin(x)/x) · (1/cos(x)). In simple terms, tangent also approximates x for small angles.
The master technique: matching sin(u)/u
Nearly every trig limit at zero can be rewritten so that sin(something) is divided by that same something. The key move is to multiply and divide by whatever makes the argument of sin match the denominator.
Example: lim(x→0) 5x / sin(2x)
Rewrite as: 5x / sin(2x) = (5/2) · (2x / sin(2x)).
As x → 0, let u = 2x → 0, and u/sin(u) → 1. So the limit is 5/2.
Example: lim(x→0) tan(x) / (2x)
Rewrite as: (1/2) · (tan(x)/x) = (1/2) · (sin(x)/x) · (1/cos(x)).
As x → 0: sin(x)/x → 1 and 1/cos(x) → 1. So the limit is 1/2.
Example: lim(x→0) sin(5x) / sin(2x)
Rewrite as: [sin(5x)/(5x)] · [5x/(2x)] · [2x/sin(2x)] = [sin(5x)/(5x)] · (5/2) · [2x/sin(2x)].
Each bracketed expression → 1. So the limit is 5/2.
The shortcut: lim(x→0) sin(ax)/sin(bx) = a/b. This works because each sin(kx) ≈ kx near zero.
Example: lim(y→49) sin(√y − 7) / (y − 49)
The trick is recognising that y − 49 = (√y − 7)(√y + 7). So:
sin(√y − 7) / (y − 49) = sin(√y − 7) / [(√y − 7)(√y + 7)]
Let u = √y − 7. As y → 49, u → 0. The expression becomes [sin(u)/u] · [1/(√y + 7)].
As u → 0: sin(u)/u → 1, and √y + 7 → 14. So the limit is 1/14.
Example: lim(s→0) sin(s² − 8s) / s
Factor the argument: s² − 8s = s(s − 8). So:
sin(s(s − 8)) / s = [sin(s(s − 8)) / (s(s − 8))] · (s − 8)
As s → 0: the bracketed ratio → 1, and (s − 8) → −8. So the limit is −8.
Example: lim(y→0) sin(5 sin y) / y
Rewrite as: [sin(5 sin y) / (5 sin y)] · [5 sin y / y] = [sin(5 sin y) / (5 sin y)] · 5 · [sin y / y].
As y → 0: sin y → 0, so 5 sin y → 0, and sin(u)/u → 1 for each. Also sin(y)/y → 1. The limit is 5.
Example: lim(y→0⁺) sin(3y) / sin(√y)
Rewrite as: [sin(3y)/(3y)] · (3y) · [√y / sin(√y)] · (1/√y).
Simplify the middle: (3y) · (1/√y) = 3√y. As y → 0⁺: 3√y → 0. So the limit is 1 · 3√y · 1 → 0.
Alternatively: sin(3y) ≈ 3y and sin(√y) ≈ √y for small y, so the ratio ≈ 3y/√y = 3√y → 0.
Limit | Value | When to use |
|---|---|---|
lim(x→0) sin(x)/x | 1 | The base case; every other limit reduces to this |
lim(x→0) x/sin(x) | 1 | The reciprocal; same result |
lim(x→0) tan(x)/x | 1 | Because tan = sin/cos and cos(0) = 1 |
lim(x→0) (1 − cos(x))/x | 0 | Cosine departs from 1 quadratically |
lim(x→0) sin(ax)/sin(bx) | a/b | Shortcut for ratio-of-sines problems |
lim(x→0) sin(ax)/(bx) | a/b | Generalised form of the fundamental limit |
The golden rule: these only work when the argument of sin (or tan) and the denominator go to zero together. If they do not, rewrite until they do.
Substitution (change of variable)
When the limit is not at zero, introduce a new variable that goes to zero. In lim(y→49) sin(√y − 7)/(y − 49), let u = √y − 7. Factor y − 49 = (√y − 7)(√y + 7) so the substitution creates a sin(u)/u form.
The general move: if the argument of sin is some expression that → 0, call that expression u and rewrite everything in terms of u.
Factoring the argument of sin
In lim(s→0) sin(s² − 8s)/s, the argument s² − 8s = s(s − 8) contains s as a factor. Dividing sin(s(s − 8)) by s(s − 8) creates the standard form, with (s − 8) left over as a separate factor that simply evaluates to −8.
Nested trig functions
In lim(y→0) sin(5 sin y)/y, the argument of the outer sin is 5 sin y. As y → 0, 5 sin y → 0 too. So you can apply sin(u)/u twice: once for the outer sin, once for the inner sin(y)/y. Multiply and divide carefully to create both standard forms.
The small-angle shortcut
For quick mental checks: near zero, sin(kx) ≈ kx and tan(kx) ≈ kx. Replace sin with its argument, simplify, and you have the limit. This is not a proof, but it gives the right answer every time on these problems, and it is a reliable way to check your algebra.
The limit sin(x)/x = 1 is why the small-angle approximation works in physics and engineering. When a pendulum swings through a small arc, sin(θ) ≈ θ lets you replace the nonlinear equation of motion with a linear one, giving the familiar period formula T = 2π√(L/g). In optics, Snell's law simplifies to n₁θ₁ = n₂θ₂ for small angles, making lens design calculations tractable. Signal processing relies on the sinc function, sin(x)/x, which describes the frequency content of a rectangular pulse.
Students often think lim(x→0) sin(x)/x = 0, reasoning that sin(0) = 0. The limit is 1, because both numerator and denominator approach zero (it is an indeterminate 0/0 form), and the rate at which they approach zero is identical.
Students apply lim sin(x)/x = 1 when x does not approach zero. The formula only works when the argument of sin and the denominator both approach zero. lim(x→5) sin(x)/x is just sin(5)/5, by direct substitution.
Students forget that the argument of sin and the denominator must match. lim(x→0) sin(3x)/x is not 1; it is 3. You must account for the coefficient.
Students use degrees instead of radians. The limit sin(x)/x = 1 requires x in radians. In degrees, lim(x→0) sin(x°)/x = π/180, which is a completely different number.
⚠️ Trig limit problems appear on nearly every Calculus I exam. The technique is always the same: rewrite to expose sin(u)/u, evaluate. Speed matters here.
⚠️ The substitution technique (for limits not at zero, like the √y − 7 problem) is tested to see whether you can adapt the standard method to non-standard setups.
⚠️ Expect at least one problem where the argument of sin needs to be factored before the standard form appears (like sin(s² − 8s)/s).
⚠️ The shortcut lim sin(ax)/sin(bx) = a/b is worth memorising. It saves time and reduces errors.
⚠️ Make sure you can handle nested trig limits (sin of sin). These look intimidating but follow the same logic applied twice.
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) = ______.
1
True or false: lim(x→0) sin(4x)/sin(9x) = 4/9.
True. The ratio of the coefficients.
Fill in the blank: lim(x→0) tan(x)/x = ______.
1
True or false: You can use lim sin(x)/x = 1 when x → 3.
False. The limit only applies as x → 0.
Q: Evaluate lim(x→0) sin(8x) / (3x).
A: Rewrite as (8/3) · sin(8x)/(8x). As x → 0, sin(8x)/(8x) → 1. The limit is 8/3.
Q: Evaluate lim(x→0) (1 − cos(x)) / (x²).
A: This is a standard result: the limit equals 1/2. It can be derived by multiplying numerator and denominator by (1 + cos(x)) and using sin²(x) = 1 − cos²(x).
Q: Evaluate lim(t→0) sin(t²) / t.
A: Rewrite as [sin(t²)/t²] · t. As t → 0, sin(t²)/t² → 1, and the extra factor of t → 0. The limit is 0.
Q: Evaluate lim(x→0) sin(3x) / sin(7x).
A: Using the shortcut: the limit is 3/7.
Q: Evaluate lim(h→0) cos(h) / h. Is this an indeterminate form?
A: No. cos(0) = 1, and 1/h → ±∞ as h → 0. This limit does not exist. The sin(x)/x technique does not apply here because the numerator does not approach zero.
Q: Evaluate lim(x→0) x · cot(x).
A: Rewrite as x · cos(x)/sin(x) = (x/sin(x)) · cos(x). As x → 0: x/sin(x) → 1 and cos(x) → 1. The limit is 1.
These limits are the proof machinery behind the trig derivative formulas. The derivative of sin(x) is derived by evaluating lim(h→0) [sin(x+h) − sin(x)]/h, which, after expanding with the angle addition formula, requires both lim sin(h)/h = 1 and lim (cos(h) − 1)/h = 0.
L'Hopital's rule (Chapter 4) provides an alternative way to evaluate these same limits. Once you learn it, you can verify your answers: the derivative of sin(2x) is 2cos(2x) and the derivative of 2x is 2, so lim sin(2x)/(2x) = 2cos(0)/2 = 1.
The Taylor series expansion of sin(x) = x − x³/3! + x⁵/5! − ... makes it visually clear why sin(x)/x → 1: dividing by x gives 1 − x²/3! + ..., and every term after the 1 vanishes as x → 0.
Trigonometric limits, sin(x)/x limit, squeeze theorem trig, fundamental trig limit, small angle approximation, sinc function, limit of tanx over x, MTH 132, Calculus I, Michigan State, section 2.4, trig limit techniques, substitution trig limits, nested sine limit, indeterminate form 0/0, special trig limits, limit sin(ax)/sin(bx), cos limit, trig limit evaluation