Why does $\lim_{x\rightarrow 0}\frac{\sin(x)}x=1$? 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. 


*

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

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

*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).

*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 $\frac{1}{2}(\text{base} \times \text{height})$.

*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?


Here is my question, how does this "section" of the circle equal to $\dfrac\theta2$ ? $($It looks like a pizza slice$)$

Well, to an arc length of $2\pi$ corresponds an area of $\pi=\dfrac{2\pi}2$ $($the full circle$)$. To an arc length of $\pi$ corresponds an area of $\dfrac\pi2$ $($the half-circle$)$. To an arc length of $\dfrac\pi2$ corresponds an area of $\dfrac\pi4$ $($the quarter-circle$)$. Etc. In general, to an arc length of $\alpha$ corresponds an area of $\dfrac\alpha2$. This follows from the symmetry of the geometric shape.
