\begin{equation} \label{tan} x<\tan x. \end{equation} Multiplying both sides of this inequality by $\cos x$ yields

\begin{equation} \label{xcos} x\cos x<\sin x. \end{equation} By this theorem,

\begin{equation} \label{x} \sin x<x. \end{equation} Combining inequalities () and () gives

\begin{equation} \label{sin} x\cos x<\sin x<x. \end{equation} Dividing by $x$ yields

\begin{equation} \label{squeeze} \cos x<\frac{\sin x}{x}<1. \end{equation} Now let $\displaystyle \frac{-\pi}{2}<x<0$ . Then $\displaystyle 0<-x<\frac{\pi}{2}$ . Plugging $-x$ into inequality () gives

\begin{equation} \label{squeeze-} \cos (-x)<\frac{\sin (-x)}{-x}<1. \end{equation} Since $\cos$ is an even function and $\sin$ is an odd function, we have

\begin{equation} \label{-squeeze} \cos x<\frac{-\sin x}{-x}<1. \end{equation} Therefore, inequality () holds for all real $x$ with $\displaystyle 0<|x|<\frac{\pi}{2}$ .

Since $\cos$ is continuous, $\displaystyle \lim_{x \to 0} \cos x=\cos 0=1$ . Thus,

\begin{equation} \label{limits} 1=\lim_{x \to 0} \cos x \le \lim_{x \to 0} \frac{\sin x}{x} \le \lim_{x \to 0} 1=1. \end{equation} By the squeeze theorem, it follows that $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$ .