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The sine-to-angle limit at the origin

Physical scope

Pure real-analysis limit used to identify the impact parameter from the asymptotic angular gap; it is not a physical assumption.

Statement

The limit over nonzero real $x$ is

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

Facts & Assumptions

Given: The real sine and cosine functions.

[F1]

Sine and cosine are differentiable on $R$ , $\sin 0 = 0$ , $\cos 0 = 1$ , and $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ (The derivatives of sine and cosine are cosine and minus sine).

[F2]

Differentiability at every point of an interval implies continuity there (A function differentiable at $c$ is continuous at $c$ ).

[F3]

The mean value theorem applies to a function continuous on a closed interval and differentiable on its interior (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

technique · apply the mean value theorem on the interval between $0$ and $x$, then use continuity of cosine at $0$

1.1givenF1F2F3algebra

Fix $x \neq 0$ . By [F1] and [F2], sine is continuous on the closed interval with endpoints $0, x$ and differentiable in its interior. By [F3] there is $ξ_{x}$ strictly between $0$ and $x$ such that $\sin x / x = (\sin x - \sin 0) / (x - 0) = \cos ξ_{x}$ .

2.1step 1.1F1F2givenalgebra∎

Given $ε > 0$ , [F1] and [F2] make cosine continuous at $0$ , so choose $δ > 0$ such that $∣ t ∣ < δ$ implies $∣ \cos t - 1 ∣ < ε$ . If $0 < ∣ x ∣ < δ$ , then $∣ ξ_{x} ∣ < ∣ x ∣ < δ$ and step 1.1 yields $∣ \sin x / x - 1 ∣ = ∣ \cos ξ_{x} - 1 ∣ < ε$ . This is the stated limit.

Depends on

Dependency tree · 2 levels

Sources

OpenStax, Calculus Volume 1, §2.3 The Limit Laws, unit-circle squeeze proof of the fundamental sine limit