# How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

How can one prove the statement
$$\lim_{x\to 0}\frac{\sin x}x=1$$
without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.

This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can’t find out how. Any help is appreciated. The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get
$$\frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1}$$
Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get
$$\cos(x)\le\frac{\sin(x)}{x}\le1\tag{2}$$
Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that
$$\lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3}$$
Also, dividing $(2)$ by $\cos(x)$, we get that
$$1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4}$$
Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that
$$\lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5}$$