Areal:\(\displaystyle\frac{r\cdot\tan x}{2}=\frac{\tan x}{2}\)
Areal:\(\displaystyle\frac{x}{2\pi r}\cdot \pi r^2 = \frac{x}{2}\)
$$x\le\tan x$$
Hans Georg Schaathun
Mars 2016
Velg→eller↓
Areal:\(\displaystyle\frac{r\cdot\tan x}{2}=\frac{\tan x}{2}\)
Areal:\(\displaystyle\frac{x}{2\pi r}\cdot \pi r^2 = \frac{x}{2}\)
$$x\le\tan x$$
$$\lim_{x\to0} \frac{\sin x}{x}$$
$$\sin x \le x \le \tan x$$
$$\frac{\sin x}{\sin x} \ge \frac{\sin x}{x} \ge \frac{\sin x}{\tan x}$$
$$1 \ge \frac{\sin x}{x} \ge \cos x$$
$$\lim_{x\to0}1 \ge \lim_{x\to0}\frac{\sin x}{x} \ge \lim_{x\to0}\cos x$$
$$1 \ge \lim_{x\to0}\frac{\sin x}{x} \ge 1$$
$$\lim_{x\to0^+} \frac{\sin x}{x}=1$$
$$x\le0$$