Limits of Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent appear throughout mathematics and physics, but evaluating their limits—especially at points where direct substitution fails—requires special techniques and fundamental identities. The concept of a trigonometric limit bridges geometry and calculus, allowing us to understand the instantaneous behavior of oscillating functions. The two most important standard trigonometric limits are lim(x→0) sin(x)/x = 1 and lim(x→0) (1 − cos(x))/x = 0. These results, which cannot be obtained simply by plugging in x = 0, form the foundation for differentiating sine and cosine from first principles. Mastering these limits unlocks the derivatives of all trigonometric functions. In this article you will learn how to evaluate trigonometric limits using the squeeze theorem, algebraic manipulation, and standard limit identities. Understanding these techniques is essential for success in calculus and for any application—from modeling wave motion to analyzing alternating electrical currents—where trigonometric functions describe change.

Key takeaways

Direct Substitution (When Possible)
The Fundamental Limit: sin(x)/x as x→0
The Second Fundamental Limit: (1 − cos x)/x as x→0
Algebraic Manipulation and Substitution
Using Trigonometric Identities
Limits at Other Points and Infinity

Direct Substitution (When Possible)

If the function is continuous at the point in question, simply substitute the value. For example, lim(x→π/3) sin(x) = sin(π/3) = √3/2, because sine is continuous everywhere. Always try direct substitution first before applying more advanced techniques.

The Fundamental Limit: sin(x)/x as x→0

Direct substitution gives 0/0 at x = 0, an indeterminate form. The geometric proof uses the squeeze (sandwich) theorem. Consider a unit circle: for 0 < x < π/2, the areas satisfy (1/2)sin(x) ≤ (1/2)x ≤ (1/2)tan(x). Dividing by (1/2)sin(x) and taking reciprocals gives cos(x) ≤ sin(x)/x ≤ 1. Since lim(x→0) cos(x) = 1, by the squeeze theorem lim(x→0) sin(x)/x = 1. This result holds for x measured in radians—never degrees.

The Second Fundamental Limit: (1 − cos x)/x as x→0

Multiply numerator and denominator by (1 + cos x): (1 − cos x)/x · (1 + cos x)/(1 + cos x) = sin²(x) / [x(1 + cos x)]. Split this as [sin(x)/x] · [sin(x)/(1 + cos x)]. As x→0, [sin(x)/x]→1 and [sin(x)/(1 + cos x)]→0/(1+1) = 0. Therefore lim(x→0)(1 − cos x)/x = 0.

Algebraic Manipulation and Substitution

Many trigonometric limits reduce to the two fundamental forms after algebraic manipulation. For instance, lim(x→0) sin(5x)/x can be rewritten as 5 · [sin(5x)/(5x)]. Let u = 5x; as x→0, u→0, so the limit becomes 5 · lim(u→0) sin(u)/u = 5 · 1 = 5. Scaling arguments of this kind are extremely common on exams.

Using Trigonometric Identities

Sometimes you need to apply identities before a limit becomes recognizable. For example, lim(x→0) sin(2x)/sin(3x) can be written as [sin(2x)/(2x)] · (2x) / { [sin(3x)/(3x)] · (3x) } = (2/3) · [sin(2x)/(2x)] / [sin(3x)/(3x)] → (2/3) · 1/1 = 2/3. Another useful identity is the double-angle formula sin(2x) = 2 sin(x)cos(x), which can simplify otherwise intractable expressions.

Limits at Other Points and Infinity

For limits as x→∞, sine and cosine do not converge because they oscillate. However, expressions like lim(x→∞) sin(x)/x = 0 can be evaluated using the squeeze theorem: since −1 ≤ sin(x) ≤ 1, we have −1/x ≤ sin(x)/x ≤ 1/x, and both bounds tend to 0. Similarly, lim(x→0⁺) x · cos(1/x) = 0 by squeezing between −x and x.

The Key Equation

\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \text{and} \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0

x — angle (radians): The independent variable representing an angle measured in radians. Both limits fail if x is in degrees because the radian measure is essential to the geometric area argument in the proof.
\sin x — sine function (dimensionless): The vertical coordinate of the point on the unit circle at angle x. For small x in radians, sin(x) ≈ x, which is why the ratio approaches 1.
1 - \cos x — versine (dimensionless): The horizontal distance from the point (cos x, sin x) to the point (1, 0) on the unit circle. For small x, this quantity behaves like x²/2, so dividing by x gives something that goes to 0.

Worked Example

Problem. Evaluate the limit: lim(x→0) [tan(3x)] / [sin(5x)]

\dfrac{3}{5}

Answer. 3/5

Figure

Diagram

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Swiss

Euler standardized the modern notation for trigonometric functions and proved countless identities linking them. His formula e^(ix) = cos x + i sin x unifies exponential and trigonometric functions, and his systematic treatment of infinite series and limits involving trigonometric expressions laid indispensable groundwork for the rigorous calculus we use today, including the fundamental limits sin(x)/x → 1. Euler lost sight in his right eye in 1738 and became almost totally blind by 1771, yet he produced roughly half of his total mathematical output after losing his vision—dictating entire papers from memory.

The Squeeze Theorem — The proof that sin(x)/x → 1 relies entirely on the squeeze theorem, which states that if f(x) ≤ g(x) ≤ h(x) near a point and the outer limits are equal, then the middle limit must equal them too.
Derivatives of Trigonometric Functions — The derivative of sin(x) is cos(x) precisely because lim(h→0) sin(h)/h = 1. Without mastering this trigonometric limit, deriving d/dx[sin x] from first principles is impossible.
Indeterminate Forms and L'Hôpital's Rule — sin(x)/x at x = 0 is the archetypal 0/0 indeterminate form. Understanding why substitution fails here motivates the study of L'Hôpital's Rule and other techniques for resolving indeterminate forms.
The Unit Circle and Radian Measure — The fundamental limit sin(x)/x = 1 is only valid when x is in radians. The entire geometric area argument depends on using radian measure, so a solid understanding of radians is a prerequisite.