Answer
\[\begin{gathered}
= 7 \hfill \\
\hfill \\
\end{gathered} \]
Work Step by Step
\[\begin{gathered}
\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\frac{{\tan \,\,7x}}{{\sin x}} \hfill \\
\hfill \\
use\,\,the\,\,identity\,\,\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} \hfill \\
\,\,\tan \,\,7x = \frac{{\sin 7x}}{{\cos 7x}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \mathop {\lim }\limits_{x \to \,0} \,\frac{{\frac{{\sin 7x}}{{\cos 7x}}}}{{\sin x}}\, = \,\,\mathop {\lim }\limits_{x \to \,0} \,\,\left( {\frac{{\sin 7x}}{{\cos 7x}}} \right)\,\left( {\frac{1}{{\sin x}}} \right) \hfill \\
\hfill \\
use\,\,the\,\,product\,\,of\,\,limits\,law \hfill \\
\hfill \\
= \,\left( {\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\frac{1}{{\cos 7x}}} \right)\,\left( {\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\,\sin 7x} \right)\,\left( {\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\frac{1}{{\sin x}}} \right) \hfill \\
\hfill \\
= \,\left( {\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\,\frac{1}{{\cos 7x}}} \right)\,\left( 7 \right)\left( {\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\,\frac{{\sin 7x}}{{7x}}} \right)\,\left( {\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\,\frac{x}{{\sin x}}} \right) \hfill \\
\hfill \\
from\,\,the\,\,theorem\,\,7.11 \hfill \\
\hfill \\
\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\,\frac{{\sin 7x}}{{7x}} = 1{\text{ and }}\mathop {\lim }\limits_{x \to \,0} \,\,\,\,\,\frac{x}{{\sin x}} = 1 \hfill \\
\hfill \\
\hfill \\
= \,\left( 1 \right)\,\left( 7 \right)\,\left( 1 \right)\,\left( 1 \right) \hfill \\
\hfill \\
= 7 \hfill \\
\end{gathered} \]
