#### Answer

=5

#### Work Step by Step

\[\begin{gathered}
\mathop {\lim }\limits_{x \to 0} \,\,\frac{{\,\,\tan \,5x}}{x} \hfill \\
\hfill \\
use\,\,the\,\,identity\,\,\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} \hfill \\
\hfill \\
\tan \,5x = \frac{{\sin 5x}}{{\cos 5x}}\,, \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\mathop {\lim }\limits_{x \to 0} \,\,\frac{{\,\,\tan \,5x}}{x} = \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin 5x}}{{x\cos 5x}} \hfill \\
\hfill \\
Use\,\,product\,\,of\,\,\,limits\, \hfill \\
\hfill \\
= 5\,\left( {\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x}}} \right)\,\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x}}{{5x}}} \right) \hfill \\
\hfill \\
where\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\sin \,5x}}{{5x}} = 1\, \hfill \\
\hfill \\
= 5\,\left( 1 \right)\,\left( 1 \right) \hfill \\
\hfill \\
= 5 \hfill \\
\end{gathered} \]