calculus - Why does $lim_{xrightarrow 0}frac{sin(x)}x=1$?

I am learning about the derivative function of $\frac{d}{dx}[\sin(x)] = \cos(x)$.

The proof stated: From $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$...

I realized I don't know why, so I wanted to learn why part is true first before moving on. But unfortunately I don't have the complete note for this proof.

1. It started with a unit circle, and then drew a triangle at $(1, \tan(\theta))$


2. It show the area of the big triangle is $\frac{\tan\theta}{2}$


3. It show the area is greater than the sector, which is $\frac{\theta}{2}$
   Here is my question, how does this "section" of the circle equal to $\frac{\theta}{2}$? (It looks like a pizza slice).



4. From there, it stated the area of the smaller triangle is $\frac{\sin(\theta)}{2}$. I understand this part. Since the area of the triangle is $\frac{1}{2}(\text{base} \times \text{height})$.




5. Then they multiply each expression by $\frac{2}{\sin(\theta){}}$ to get
   $\frac{1}{\cos(\theta)} \ge \frac{\theta}{\sin(\theta)} \ge 1$

And the incomplete notes ended here, I am not sure how the teacher go the conclusion $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. I thought it might be something to do with reversing the inequality... Is the answer obvious from this point? And how does step #3 calculation works?

Answer

Draw the circle of radius $1$ centered at $(0,0)$ in the Cartesian plane.

Let $\theta$ be the length of the arc from $(1,0)$ to a point on the circle. The radian measure of the corresponding angle is $\theta$ and the height of the endpoint of the arc above the coordinate axis is $\sin\theta$.

Now look at what happens when $\theta$ is infinitesimally small. The length of the arc is $\theta$ and the height is also $\theta$, since that infinitely small part of the circle looks like a vertical line (you're looking at the neighborhood of $(1,0)$ under a microscope).

Since $\theta$ and $\sin\theta$ are the same when $\theta$ is infinitesimally small, it follows that $\dfrac{\sin\theta}\theta=1$ when $\theta$ is infinitesimally small.

That is how Leonhard Euler viewed the matter in the 18th century.

Why does the sector of the circle have area $\theta/2$?

The whole circle has area $\pi r^2=\pi 1^2 = \pi$. The fraction of the circle in the sector is
$$\frac{\text{arc}}{\text{circumference}} = \frac{\theta}{2\pi}.$$
So the area is
$$\frac \theta {2\pi}\cdot \pi = \frac\theta2.$$