Fundamental trigonometric limit

Take $0<x<\frac{\pi}{2}$ on the unit circle with center $O$, point $A=(1,0)$, point $B=(\cos x,\sin x)$ on the circle, and point $T=(1,\tan x)$ where the ray $OB$ meets the vertical tangent at $A$.

The triangle $\triangle OAB$ sits strictly inside the circular sector $OAB$, which sits strictly inside the right triangle $\triangle OAT$. Comparing their areas gives $\frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\tan x$.

Since $\sin x>0$ on $(0,\frac{\pi}{2})$, multiply through by $\frac{2}{\sin x}$: $1 < \frac{x}{\sin x} < \frac{1}{\cos x}$. Taking reciprocals reverses the inequalities: $\cos x < \frac{\sin x}{x} < 1$.

As $x\to 0^+$, $\cos x\to 1$, so the Squeeze Theorem forces $\lim_{x\to 0^+}\frac{\sin x}{x}=1$. Finally, $g(x)=\frac{\sin x}{x}$ is an even function ($g(-x)=\frac{\sin(-x)}{-x}=\frac{-\sin x}{-x}=g(x)$), so $\lim_{x\to 0^-}\frac{\sin x}{x}=1$ as well, giving $\lim_{x\to 0}\frac{\sin x}{x}=1$. $\blacksquare$