Limit of sin(x)/x as x → 0 Equals 1: Three Proofs

The limit of sinx/x as x approaches 0 is 1. This holds when x is in radians. Direct substitution gives the indeterminate form 0/0. We establish this fundamental limit geometrically. The squeeze theorem, power series, or formal definitions prove it.

When you evaluate this limit on the unit circle, an angle of x radians subtends an arc of length x. The vertical height of the corresponding point on the circle is sin x. As the angle shrinks toward 0, the curved arc and the vertical segment become nearly identical in length, driving their ratio toward 1. Graphically, this means the line y = x is tangent to y = sin x at the origin, a property formalised by linear approximation as sin x ≈ x for small angles. Physicists rely on this small-angle approximation in simple pendulum equations and wave optics.

Comparing the three proof methods

While all three approaches establish that the limit is 1, they differ in prerequisites, mathematical rigor, and risk of circular reasoning.

1. The geometric squeeze theorem proof

The standard, non-circular proof uses geometry on the unit circle to bound sin x / x between two expressions that both approach 1.

Consider an angle x in radians such that 0 < x < π/2. Place the vertex of the angle at the origin O(0,0) on a unit circle, with one side along the positive x-axis passing through point A(1,0), and the other side intersecting the circle at point P(cos x, sin x). This setup is fundamental for understanding concepts like the limit of sinx/x. Next, construct a line perpendicular to the x-axis at A, extending the ray OP to intersect this line at point T(1, tan x).

Comparing three nested geometric regions sharing the vertex at O:

1. Inner triangle (OAP): Base is 1 and height is sin x, so its area is (1/2) sin x.
2. Circular sector (OAP): With central angle x radians on a unit circle, its area is (1/2) x.
3. Outer triangle (OAT): Base is 1 and height is tan x, so its area is (1/2) tan x.

Because each region completely contains the preceding one, their areas follow a strict order:

$$\frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\tan x$$

Divide every term by (1/2) sin x, which is strictly positive for 0 < x < π/2:

$$1 < \frac{x}{\sin x} < \frac{1}{\cos x}$$

Take the reciprocal of all terms, which flips the inequality signs:

$$\cos x < \frac{\sin x}{x} < 1$$

Now take the limit as x approaches 0 from the right (x → 0⁺):

$$\lim_{x \to 0^+} \cos x = 1 \quad \text{and} \quad \lim_{x \to 0^+} 1 = 1$$

By the squeeze theorem, the right-hand limit must also be 1. To handle negative angles (x → 0⁻), note that f(x) = sin(x)/x is an even function, because sin(-x)/(-x) = (-sin x)/(-x) = sin(x)/x. Since the left-hand and right-hand limits are equal, the two-sided limit of sinx/x exists and equals 1.

2. The Taylor series expansion proof

If sine is defined analytically through power series rather than geometric ratios, you can evaluate the limit algebraically without geometry.

The Maclaurin series for sin x is defined for all real numbers as:

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$

For any non-zero value of x, divide the power series term-by-term by x using Taylor series principles:

$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots$$

Taking the limit as x → 0 renders every term containing x equal to 0:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1 - 0 + 0 - 0 + \cdots = 1$$

This proof is mathematically rigorous provided the definition of sine does not depend on unit circle geometry. However, if sine was defined geometrically, deriving its Taylor series coefficients requires knowing that the derivative of sine is cosine, which itself relies on the limit of sinx/x.

3. L’Hôpital’s rule (and why it can be circular)

Applying L’Hôpital’s rule offers a fast algebraic check for indeterminate forms of type 0/0.

Direct substitution into lim (x → 0) sin x / x yields 0/0. Differentiating numerator and denominator with respect to x gives:

$$\lim_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\cos x}{1} = \frac{\cos 0}{1} = 1$$

Warning regarding circular logic: In standard calculus curricula, proving that d/dx (sin x) = cos x requires evaluating lim (h → 0) sin(h)/h from the difference quotient definition. Using L’Hôpital’s rule to establish the limit assumes the derivative you are trying to prove. However, it is only non-circular if trigonometric functions are defined independently via differential equations or complex numbers.

4. Numerical table and rate of convergence

Evaluating the function at numbers progressively closer to 0 provides clear numerical evidence for the limit value.

The output approaches 1 symmetrically from below for both positive and negative inputs. Notice that each 10-fold decrease in x reduces the difference from 1 by a factor of roughly 100. This quadratic convergence rate matches the second term of the Taylor expansion, where the error gap is approximately x² / 6.

Why this limit matters: the derivative of sine

This limit, the limit of sinx/x, serves as the foundational building block for all trigonometric calculus. Also, using the limit definition of the derivative and the sine angle-addition identity:

$$\begin{aligned}\frac{d}{dx}\sin x &= \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h} &= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} &= \sin x \cdot \lim_{h \to 0} \frac{\cos h - 1}{h} + \cos x \cdot \lim_{h \to 0} \frac{\sin h}{h}\end{aligned}$$

The first secondary limit equals 0, while the second secondary limit is our central limit, which equals 1. Substituting these values yields:

$$\frac{d}{dx}\sin x = \sin x \cdot (0) + \cos x \cdot (1) = \cos x$$

Every subsequent calculus derivative involving trigonometric expressions—such as the derivative of tan x—relies on this exact result.

Why radians are required in calculus

If you evaluate lim (x → 0) sin(x°) / x using degree measure, the limit is not 1. Converting x degrees to radians, for example for the limit of sinx/x, introduces a scaling factor of π / 180:

$$\lim_{x \to 0} \frac{\sin(x^\circ)}{x} = \lim_{x \to 0} \frac{\sin(\frac{\pi x}{180})}{x} = \frac{\pi}{180} \approx 0.0174533$$

Using degrees in calculus would introduce an inconvenient factor of π / 180 into every derivative formula. Radians, however, keep the derivative of sine purely as cosine without extra constant multipliers.

Worked examples: understanding the limit of sinx/x

To evaluate more complex trigonometric limits, transform the expression until the argument inside the sine function matches the denominator exactly.

Example 1. Evaluate lim (x → 0) sin(5x) / x. Multiply numerator and denominator by 5:

$$\frac{\sin 5x}{x} = 5 \cdot \frac{\sin 5x}{5x} \to 5 \cdot 1 = 5$$

Example 2. Evaluate lim (x → 0) sin(3x) / sin(2x). Multiply and divide by 3x and 2x:

$$\frac{\sin 3x}{\sin 2x} = \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{3x}{2x} \to 1 \cdot 1 \cdot \frac{3}{2} = \frac{3}{2}$$

Example 3. Evaluate lim (x → 0) tan(x) / x. Also, rewrite tangent as sine over cosine:

$$\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x} \to 1 \cdot 1 = 1$$

Example 4. Evaluate lim (x → 0) (1 – cos x) / x². Multiply by the conjugate (1 + cos x):

$$\frac{1 - \cos x}{x^2} = \frac{1 - \cos^2 x}{x^2(1 + \cos x)} = \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x} \to 1^2 \cdot \frac{1}{2} = \frac{1}{2}$$

If the denominator is x instead of x², the extra factor of x forces the limit to 0.

Example 5. Evaluate lim (x → 0) sin(x²) / x. Rewrite as x · (sin(x²) / x²):

$$\lim_{x \to 0} \left( x \cdot \frac{\sin(x^2)}{x^2} \right) = 0 \cdot 1 = 0$$

Example 6. Evaluate lim (x → π) sin(x) / (π – x). Substitute u = π – x, meaning sin x = sin(π – u) = sin u. As x → π, u → 0:

$$\lim_{u \to 0} \frac{\sin u}{u} = 1$$

Example 7. Evaluate lim (x → 0) sin(2x) / (x cos x). Factor into manageable limit terms:

$$\frac{\sin 2x}{x \cos x} = 2 \cdot \frac{\sin 2x}{2x} \cdot \frac{1}{\cos x} \to 2 \cdot 1 \cdot 1 = 2$$

Common mistakes to avoid

• Calculator set to degrees. Evaluating trigonometric ratios on a calculator set to degree mode gives ~0.01745 instead of 1.
• Incorrect algebraic cancellation. Treating sin(5x) / sin(2x) as if sine were a variable that cancels out (giving 5/2) is algebraically invalid, even though the numerical limit matches.
• Confusing x → 0 with x → ∞. As x → ∞, sin(x) / x approaches 0 because the numerator remains bounded between -1 and 1 while the denominator grows infinitely. Refer to limits at infinity.
• Assuming all argument coefficients yield 1. The argument inside sine must match the denominator exactly; for example, sin(3x) / x approaches 3, not 1.

Interactive limit solver

You can use the limit calculator to evaluate one-sided limits or limits at infinity, or check your steps using the L’Hôpital’s rule calculator.

Practice problems

Test your understanding before reviewing the solutions below.

1. $$\lim_{x \to 0} \frac{\sin 7x}{3x}$$
2. $$\lim_{x \to 0} \frac{x}{\sin x}$$
3. $$\lim_{x \to 0} \frac{\sin^2 x}{x^2}$$
4. $$\lim_{x \to 0} \frac{\tan 3x}{x}$$
5. $$\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}$$
6. $$\lim_{x \to 0} \frac{\tan 3x}{\sin 5x}$$

Answers: (1) 7/3; (2) 1; (3) 1; (4) 3; (5) 2 (since 1 – cos(2x) = 2 sin²(x)); (6) 3/5.

Frequently asked questions

Is sin(x)/x defined at x = 0?

No, the function is undefined at x = 0 because division by zero is undefined. However, because the limit of sinx/x as x approaches 0 exists and equals 1, defining f(0) = 1 creates a continuous function known as the unnormalized sinc function.

Why is using L’Hôpital’s rule considered circular for this limit?

L’Hôpital’s rule requires taking the derivative of sin(x), which is cos(x). In introductory calculus, proving that the derivative of sin(x) is cos(x) relies on knowing that lim (x → 0) sin(x)/x = 1, creating circular reasoning unless sine is defined non-geometrically.

What is the limit of sin(x)/x as x approaches infinity?

As x approaches infinity, the limit of sinx/x is 0. The numerator oscillates continuously between -1 and 1, while the denominator grows without bound, driving the quotient to 0.

How does this limit change if x is measured in degrees?

When x is measured in degrees, the limit of sinx/x as x approaches 0 equals π / 180, which is approximately 0.01745. Radians are used in calculus specifically because they eliminate this constant multiplier.

Further reading

• Proof of Trig Limits (Paul’s Online Math Notes) — Geometric proofs for sine and cosine limits.
• Computing Limits (Paul’s Online Math Notes) — Methods for evaluating indeterminate limits using the squeeze theorem.