Beweis: Grenzwert \(\frac{\sin(\theta)}{\theta}\)

\begin{align*} \lim_{\theta \to 0^+}\left(\frac{\sin(\theta)}{\theta}\right) =& 1 \end{align*}

• Einschnürungssatz

Skizze fehlt:

\begin{align*} A_1 =& \frac{1}{2} \cdot \sin(\theta)\cos(\theta) \\ A_2 =& r^2 \cdot \pi \cdot \frac{\theta}{2\pi} \\ =& \frac{1}{2}\theta \\ A_3 =& \frac{1}{2} \cdot 1 \cdot \tan(\theta) \\ =& \frac{1}{2}\tan(\theta) A_1 \leq A_2 \leq A_3 \\ \frac{1}{2} \sin(\theta)\cos(\theta) \leq \frac{1}{2} \theta \leq \frac{1}{2} \tan(\theta) \vert && \cdot 2\\ \sin(\theta)\cos(\theta) \leq \theta \leq \tan(\theta) && (…)^{-1}\\ \frac{1}{\sin(\theta)\cos(\theta)} \leq \frac{1}{\theta} \leq \frac{\cos(\theta)}{\sin(\theta)} \vert && \cdot \sin(\theta)\\ \frac{1}{\cos(\theta)} \leq \frac{\sin(\theta)}{\theta} \leq \cos(theta) \end{align*} \begin{align*} \lim_{\theta \to 0}\left(\frac{1}{\cos(\theta)}\right) \leq \lim_{\theta \to 0}\left(\frac{\sin(\theta)}{\theta}\right) \leq \lim_{\theta \to 0}\left(\cos(theta)\right) \\ 1 \leq \lim_{\theta \to 0}\left(\frac{\sin(\theta)}{\theta}\right) \leq 1 \\ \Rightarrow \lim_{\theta \to 0}\left(\frac{\sin(\theta)}{\theta}\right) = 1 \end{align*}