# The limit of sin(x) over x

## Statement

The limit as $\theta$ goes to $0$ of $\sin(\theta) \div \theta$ converges to $1$, for radians:

$$\lim_{\theta \to 0} \left(\frac{\sin(\theta)}{\theta}\right) = 1$$

## Proof

Use the unit circle with radius $1$ and draw a triangle $OAB$ inside it, with angle $\theta$. Then draw the line $CD$ perpendicular to the cirlce and the $x$-axis, like the image below:

Since $OB = OC = 1$, we can find the values of $AB$, $OA$ and $CD$:

$$\sin(\theta) = \frac{AB}{OB} = AB$$

$$\cos(\theta) = \frac{OA}{OB} = OA$$

$$\tan(\theta) = \frac{CD}{OC} = CD$$

Note that the area of the unit circle is $π \cdot 1^2 = π$. From this follows that the area of circle segment $OBC$ is $π \cdot \frac{\theta}{2π} = \frac{1}{2} \theta$, when $\theta$ is in radians.

Next, find the areas of $\triangle OAB$ and $\triangle OCD$.

$$A_{\triangle OAB} = \tfrac{1}{2} \cdot OA \cdot AB = \tfrac{1}{2}\sin(\theta)\cos(\theta)$$

$$A_{\triangle OCD} = \tfrac{1}{2} \cdot OC \cdot CD = \tfrac{1}{2}\tan(\theta)$$

Now note the following inequality:

$$A_{\triangle OAB} \le A_{OBC} \le A_{\triangle OCD}$$

Substitute these values and multiply each side by $2$. Note that tangent is sine divided by cosine.

$$\tfrac{1}{2}\sin(\theta)\cos(\theta) \le \tfrac{1}{2}\theta \le \tfrac{1}{2}\tan(\theta)$$

$$\sin(\theta)\cos(\theta) \le \theta \le \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$

Divide each side by $\sin(\theta)$. Since $\sin(\theta)$ is positive for a small and positive $\theta$, the inequality sign stays as is.

$$\cos(\theta) \le \frac{\theta}{\sin(\theta)} \le \frac{1}{\cos(\theta)}$$

Flip each fraction. This reverses the inequality sign.

$$\frac{1}{\cos(\theta)} \ge \frac{\sin(\theta)}{\theta} \ge \cos(\theta)$$

Now take the limit as $\theta$ approached $0$ on each side.

$$\lim_{\theta \to 0} \left(\frac{1}{\cos(\theta)}\right) \ge \lim_{\theta \to 0} \left(\frac{\sin(\theta)}{\theta}\right) \ge \lim_{\theta \to 0} \bigg(\cos(\theta)\bigg)$$

Since $\cos(0) = 1$, the first and third limit can be evaluated.

$$1 \ge \lim_{\theta \to 0} \left(\frac{\sin(\theta)}{\theta}\right) \ge 1$$

From this follows:

$$\lim_{\theta \to 0} \left(\frac{\sin(\theta)}{\theta}\right) = 1$$

## Proofs building upon this proof

### Derivative of cos(x)

This proof shows what the derivative of cos(x) is.

### Derivative of sin(x)

This proof shows what the derivative of sin(x) is.