Answer
$$\lim _{\theta \rightarrow 0} \frac{\cos \theta-1}{2 \theta^{2}}=-\frac{1}{4} $$
Work Step by Step
Given $$\lim _{\theta \rightarrow 0} \frac{\cos \theta-1}{2 \theta^{2}}$$
\begin{aligned}
\lim _{\theta \rightarrow 0} \frac{\cos \theta-1}{2 \theta^{2}} &=\lim _{\theta \rightarrow 0} \frac{\cos \theta-1}{2 \theta^{2}} \cdot \frac{\cos \theta+1}{\cos \theta+1}\\
&=\lim _{\theta \rightarrow 0} \frac{\cos ^{2} \theta-1}{2 \theta^{2}(\cos \theta+1)}=\lim _{\theta \rightarrow 0} \frac{-\sin ^{2} \theta}{2 \theta^{2}(\cos \theta+1)} \\
&=-\frac{1}{2} \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{\theta} \cdot \frac{1}{\cos \theta+1}\\
&=-\frac{1}{2} \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \cdot \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \cdot \lim _{\theta \rightarrow 0} \frac{1}{\cos \theta+1} \\
&=-\frac{1}{2} \cdot 1 \cdot 1 \cdot \frac{1}{1+1}=-\frac{1}{4} \end{aligned}
