Beweis: Grenzwert \(\frac{1-\cos(\theta)}{\theta}\)
1. Satz
\begin{align*}
\lim_{\theta \to 0^+}\left(\frac{1-\cos(\theta)}{\theta}\right) =& 0
\end{align*}
2. Lemma
- Grenzwert: \(\frac{\sin(\theta)}{\theta}\)
- trigonometrischer pythagoras
3. Beweis
\begin{align*}
\lim_{\theta \to 0^+}\left(\frac{1-\cos(\theta)}{\theta}\right) =& \lim_{\theta \to 0^+}\left(\frac{1-\cos(\theta)}{\theta} \cdot \frac{1+\cos(\theta)}{1+\cos(\theta)}\right) \\
=& \lim_{\theta \to 0^+}\left(\frac{1-\cos^2(\theta)}{\theta + \theta\cos(\theta)}\right) \\
=& \lim_{\theta \to 0^+}\left(\frac{\sin^2(\theta)}{\theta \left(1 + \cos(\theta)\right)} \right) \\
=& \lim_{\theta \to 0^+}\left(\frac{\sin(\theta)}{\theta} \cdot \frac{\sin(\theta}{\left(1 + \cos(\theta)\right)} \right) \\
=& \lim_{\theta \to 0^+}\left(\frac{\sin(\theta)}{\theta}\right) \cdot \lim_{\theta \to 0^+}\left(\frac{\sin(\theta}{\left(1 + \cos(\theta)\right)} \right) \\
=& 1 \cdot 0 \\
=& 0
\end{align*}