Doubt on a proof that $\lim_{x\rightarrow0}{\frac{\sin{x}}{x}}=1$ The full proof can be found here. Basically, we compare the three areas that depend on $x$ in the circle of radius $1$ shown below. 

Regardless of the value of $x$, we should have 
$$\text{area of sector OAC} < \text{area of triangle OAP} < \text{area of sector OBP}$$
$$\frac{1}{2}x(\cos{x})^2<\frac{1}{2}(\cos{x})(\sin{x})<\frac{x}{2}$$
I have two questions about it:
1) Shouldn't there be a $\le$ instead of the $<$ sign in the inequality above, since the areas of sector $OAC$ and the area of triangle $OAP$ both become zero when $x=\frac{\pi}{2}$?
2) If the value of $x$ is such that we end up in the fourth quadrant, the value of $\sin{x}$ becomes negative and the inequality no longer holds (since $\frac{1}{2}x(\cos{x})^2>0$ and $\frac{1}{2}(\cos{x})(\sin{x})<0$). How can we go around this?
Thanks in advance.

1) When we are talking about $x\to0$, we are looking at those $x$ close to $0$, but not equal to $0$. The value of the function at $x=0$ is just irrelevent. So $<$ is correct.
2) For $\frac{\pi}{2}>x>0$, we have
$$\frac{1}{2}x(\cos{x})^2<\frac{1}{2}(\cos{x})(\sin{x})<\frac{x}{2}$$
This implies that
$$\cos x<\frac{\sin x}{x}<\frac{1}{\cos x}$$
For $\frac{-\pi}{2}<x<0$, let $y=-x$. Then $\frac{\pi}{2}>y>0$ and hence 
$$\cos y<\frac{\sin y}{y}<\frac{1}{\cos y}$$
Note that $\sin x=\sin(-y)=-\sin y$ and $\cos x=\cos(-y)=\cos y$.
So we still have 
$$\cos x<\frac{\sin x}{x}<\frac{1}{\cos x}$$
