Answer
\[\begin{gathered}
= 0 \hfill \\
\hfill \\
\end{gathered} \]
Work Step by Step
\[\begin{gathered}
\mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\,\,\frac{{\sec \,\,\,\theta - 1}}{\theta } \hfill \\
\hfill \\
identity\,\,\,\,\sec \,\,\theta = \frac{1}{{\cos \,\,\theta }}\,\, \hfill \\
then \hfill \\
\hfill \\
= \mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\frac{{\frac{1}{{\cos \theta }} - 1}}{\theta } = \mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\frac{{\frac{{1 - \cos \,\theta }}{{\cos \,\theta }}}}{\theta } \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\frac{{1 - \cos \theta }}{{\theta \,\,\cos \,\,\theta }} \hfill \\
\hfill \\
= \,\left( {\mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\,\,\frac{1}{{\cos \,\,\theta }}} \right)\,\,\,\,\left( {\mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\frac{{1 - \cos \,\,\theta }}{\theta }} \right) \hfill \\
\hfill \\
from\,\,the\,\,theorem\,\,7.11\,\,\mathop {\lim }\limits_{\theta \, \to \,0} \,\,\,\frac{{1 - \cos \,\,\theta }}{\theta } = 0 \hfill \\
\hfill \\
= \,\left( 1 \right)\,\left( 0 \right) = 0 \hfill \\
\hfill \\
\end{gathered} \]
