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