$\lim_{x \to 0} \frac{\sin x}{x} = 1$
This is a unit circle, so there two radii are labeled with length $1$. The vertical line labeled $\sin{x}$ because sine of the angle $x$ is that line over the hypotenuse ($1$). The second vertical line is labeled $\tan{x}$ because tangent of the angle $x$ is that line over the base of the triangle ($1$).
Compare the areas of three regions on the diagram above:
 Blue: $\frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(1)(\sin{x}) = \frac{1}{2}\sin{x}$
 Blue + yellow: $\frac{x}{2\pi}(\pi) = \frac{x\pi}{2\pi} = \frac{x}{2}$
 The area of that sector of the circle is that part of the circle ($x$ radians out of $2\pi$ total radians around the circle) times the area of the circle ($\pi r^2$, or $\pi 1^2$, so $\pi$).
 Blue + yellow + red: $\frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(1)(\tan{x}) = \frac{1}{2}\tan{x}$
We can see visually that these areas can be arranged in order of size like this:

$\frac{1}{2} \sin{x} < \frac{x}{2} < \frac{1}{2} \tan{x}$

$\sin{x} < x < \tan{x}$

$\frac{\sin{x}}{\sin{x}} < \frac{x}{\sin{x}} < \frac{\tan{x}}{\sin{x}}$
 Divide each term by $\sin{x}$

$1 < \frac{x}{\sin{x}} < \frac{1}{\cos{x}}$

$1 > \frac{\sin{x}}{x} > \cos{x}$
 Take the reciprocal of each term, and therefore flip inequality signs
By the Squeeze Theorem, we can see that $\frac{\sin{x}}{x}$ is "squeezed" between $1$ and $\cos{x}$. As $x$ tends to $0$, $\cos{x}$ tends to $1$ as well.
Therefore, $\frac{\sin{x}}{x}$ is squeezed between 1 and 1 as x → 0, so the limit of $\frac{\sin{x}}{x}$ is 1.