Why is this nontrivial?

At $x = 0$, both the numerator $\sin x$ and the denominator $x$ equal zero. If you try direct substitution:

$$\frac{\sin 0}{0} = \frac{0}{0} \quad \text{— indeterminate form!}$$

This is a genuine $\frac{0}{0}$ indeterminate form. The expression $\frac{0}{0}$ is not a number — it could "equal" anything. A special argument is needed to determine that the limit is exactly $1$.

Geometric Intuition

On the unit circle, for a small angle $x$ (in radians):

• The arc length subtended by the angle is exactly $x$ (by definition of radian measure).
• The sine $\sin x$ is the vertical projection of the radius — the chord's $y$-component.

For small angles, the arc and the chord become nearly indistinguishable. The sine is practically equal to the angle itself. This is the geometric heart of the limit.

Notice how $\frac{\sin x}{x} \to 1$ as $x \to 0$ from both sides. The function is even: $\frac{\sin(-x)}{-x} = \frac{\sin x}{x}$.