sin x / x tends to 1

Statement

$$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$

Remarks

The quotient is the difference quotient of $\sin$ at $0$. So the limit is $\sin'(0) = \cos 0 = 1$, by the derivative of sine. The series gives $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \cdots$ for $x \ne 0$. A geometric proof uses $\cos x \le \frac{\sin x}{x} \le 1$ for $0 < |x| < \pi/2$ and the squeeze theorem. Mathlib states it as Real.tendsto_sin_div_nhdsNE_zero.

Explanations

L1 human

For small angles in radians, $\sin x$ is almost equal to $x$. The ratio of the two tends to $1$.