Answer
$$\frac{4}{9}$$
Work Step by Step
\begin{align*}
\lim _{x\rightarrow 0} \frac{ \tan 4x}{9x}&=\lim _{x\rightarrow 0} \frac{ \sin 4x}{9x} \frac{1}{\cos 4x} \\
&= \lim _{x\rightarrow 0} \frac{4}{9}\frac{ \sin 4x}{4x} \frac{1}{\cos 4x} \\
&= \frac{4}{9}\lim _{4x\rightarrow 0} \frac{ \sin 4x}{4x} \lim _{x\rightarrow 0} \frac{1}{\cos 4x}\\
&= \frac{4}{9}\frac{1}{\cos 0}\\
&=\frac{4}{9}.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
