Given that \(\lim_{\theta\rightarrow0}\frac{\cos^2\theta-1}{\theta}\)

Compute the limit as follows.

\(\lim_{\theta\rightarrow0}\frac{\cos^2\theta-1}{\theta}=\lim_{\theta\rightarrow0}\frac{-(1-\cos^2\theta)}{\theta}\)

\(=\lim_{\theta\rightarrow0}\frac{-\sin^2\theta}{\theta}\)

\(=\lim_{\theta\rightarrow0}\frac{\sin\theta}{\theta}\times\lim_{\theta\rightarrow0}(-\sin\theta)\)

\(=1\times(0)\)

\(=0\)

Thus, \(\lim_{\theta\rightarrow0}\frac{\cos^2\theta-1}{\theta}=0\)

Compute the limit as follows.

\(\lim_{\theta\rightarrow0}\frac{\cos^2\theta-1}{\theta}=\lim_{\theta\rightarrow0}\frac{-(1-\cos^2\theta)}{\theta}\)

\(=\lim_{\theta\rightarrow0}\frac{-\sin^2\theta}{\theta}\)

\(=\lim_{\theta\rightarrow0}\frac{\sin\theta}{\theta}\times\lim_{\theta\rightarrow0}(-\sin\theta)\)

\(=1\times(0)\)

\(=0\)

Thus, \(\lim_{\theta\rightarrow0}\frac{\cos^2\theta-1}{\theta}=0\)