Establish the formula $$\bbox[5px, border:1px solid black, #F9F7ED]{ \lim\limits_{\theta \rightarrow 0}\dfrac{\cos \theta -1}{\theta} = 0}$$

where $\theta$ is measured in radians.

Solution First we rewrite the expression $\dfrac{\cos \theta -1}{\theta}$ as the product of two terms whose limits are known. For $\theta \neq 0$, $$\begin{eqnarray*} \frac{\cos \theta -1}{\theta }&=&\left( \frac{\cos \theta -1}{\theta }\right)\! \left( \frac{\cos \theta +1}{\cos \theta +1}\right)\\ &=&\frac{\cos ^{2}\theta -1 }{\theta (\cos \theta +1)}\underset{\color{#0066A7}{{\sin ^{2}{\theta +}\cos ^{2} {\theta =1}}}}{\underset{\color{#0066A7}{\uparrow }}{=}}\frac{-\sin ^{2}\theta }{\theta (\cos \theta +1)} = \left( \frac{\sin \theta }{\theta }\right) \frac{(-\sin \theta)}{\cos \theta +1} \end{eqnarray*}$$

Now we find the limit. $$\begin{eqnarray*} \lim_{\theta \rightarrow 0}\frac{\cos \theta -1}{\theta } &=&\lim_{\theta \rightarrow 0}\left[ \left( \frac{\sin \theta }{\theta } \right) \!\left( \frac{(-\sin \theta )}{\cos \theta +1}\right) \right] = \left[ \lim_{\theta \rightarrow 0}\frac{ \sin \theta }{\theta }\right]~ \!\left[ \lim_{\theta \rightarrow 0} \frac{ -\sin\theta}{\cos \theta +1} \right] \nonumber \\ &=& 1\,{\cdot}\,\dfrac{\lim\limits_{\theta \rightarrow 0}(-\sin \theta)}{\lim\limits_{\theta \rightarrow 0}(\cos\theta+1)} =\dfrac{0}{2} =0 \end{eqnarray*}$$