第14題，
\(\displaystyle \lim_{\theta\to0}\frac{\tan\theta - \sin\theta}{\theta^3}=\lim_{\theta\to0}\frac{\sin\theta}{\theta^3}\cdot\frac{1-\cos\theta}{\cos\theta}\)
\(\displaystyle=\lim_{\theta\to0}\frac{\sin\theta}{\theta^3}\cdot\frac{1-\cos\theta}{\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta}=\lim_{\theta\to0}\frac{\sin^3\theta}{\theta^3}\cdot\frac{1}{\cos\theta\left(1+cos\theta\right)}\)
\(\displaystyle=1^3\cdot\frac{1}{1\cdot\left(1+1\right)}=\frac{1}{2}\)
不放心的話，頂多只需要再補上 \(\displaystyle\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1\) 的證明就可以了。
