If $\theta$ is measured in radians, then $ \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 $ ### Proof ![[Unit Circle and Tangent.png]] From the figure, it can be seen that when $0<\theta<\frac{\pi}{2}$, $\sin(\theta) \leq \theta \leq \tan(\theta)$ we can divide by $\sin(\theta)$ since it is positive on the interval $ 1 \leq \frac{\theta}{\sin(\theta)} \leq \sec(\theta)$ Taking the reciprocal of each part (and therfore changing the order) $ \cos(\theta) \leq \frac{\sin(\theta)}{\theta} \leq 1$ Knowing $\lim_{\theta \rightarrow 0^+}{\cos(\theta)}=1$ because $\cos(0)=1$ is enough to use [[Squeeze Theorem]] to say that $\lim_{\theta\rightarrow 0^+}\frac{\sin(\theta)}{\theta}=1$ The even symmetry of $\frac{\sin(\theta)}{\theta}$ means that $\lim_{\theta\rightarrow 0^-}\frac{\sin(\theta)}{\theta}=1$ Since both [[One-sided Limits]] are equal, $\lim_{\theta\rightarrow 0}\frac{\sin(\theta)}{\theta}=1$