Two special Trigonometrical limits

Result 9.1. (a) $\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ (b) $\displaystyle \lim_{\theta\to0}\frac{1-\cos\theta}{\theta}=0$.

Proof of (a), via areas on the unit circle (Figs. 9.28–9.31). Take the unit circle centred at the origin, and let $\theta$ be a small positive angle, with $P=(\cos\theta,\sin\theta)$ the corresponding point on the circle. Three regions, all built on the same angle $\theta$, can be compared by area:

• the inscribed triangle with vertices at the origin, $(1,0)$ and $P$ has area $\tfrac12\sin\theta$;
• the circular sector of angle $\theta$ has area $\tfrac12\theta$ (the standard sector-area fact, with the angle measured in radians);
• the circumscribing right triangle, obtained by erecting a vertical segment of length $\tan\theta$ at $(1,0)$, has area $\tfrac12\tan\theta$.

Since the inscribed triangle sits inside the sector, which in turn sits inside the outer triangle, their areas satisfy $\tfrac12\sin\theta\le\tfrac12\theta\le\tfrac12\tan\theta$, i.e. $\sin\theta\le\theta\le\tan\theta$. Multiplying through by the positive quantity $\dfrac2{\sin\theta}$ turns this into $1\le\dfrac{\theta}{\sin\theta}\le\dfrac1{\cos\theta}$, and taking reciprocals throughout (which reverses both inequalities) gives

Because $\cos(-\theta)=\cos\theta$ and $\dfrac{\sin(-\theta)}{-\theta}=\dfrac{\sin\theta}{\theta}$, this same chain of inequalities holds for negative $\theta$ too, so it is valid on a full punctured neighbourhood of $0$ inside $\left(-\tfrac\pi2,\tfrac\pi2\right)$. Since $\lim_{\theta\to0}\cos\theta=1$ and the constant function $1$ trivially has limit $1$, the Sandwich Theorem (Section 9.2.8) squeezes $\dfrac{\sin\theta}{\theta}$ to the common value $1$ as well. $\blacksquare$

Proof of (b), via a half-angle identity. Rewrite $1-\cos\theta$ using the identity $1-\cos\theta=2\sin^2\!\left(\dfrac\theta2\right)$, so that

$\frac{1-\cos\theta}{\theta}=\frac{2\sin^2(\theta/2)}{\theta}=\sin\!\left(\frac\theta2\right)\cdot\frac{\sin(\theta/2)}{\theta/2}.$ …