Section 4.4 Exercise 10

$\begin{align*}\lim_{x\to 0}\frac{\sin 4x}{\tan 5x} & =\lim_{x\to 0}\frac{\sin 4x\cdot \cos 5x}{\sin 5x}\\
& =\lim_{x\to 0}\frac{4}{5}\frac{\sin 4x}{4}\frac{5}{\sin 5x}\cos 5x\\
& =\frac{4}{5}\cdot \lim_{x\to 0}\left(\frac{\sin 4x}{4}\right)\cdot \lim_{x\to 0}\left(\frac{5}{\sin 5x}\right)\cdot \lim_{x\to 0}\left(\cos 5x\right) \\
& =\frac{4}{5}\cdot 1 \cdot 1 \cdot 1=\frac{4}{5}
\end{align*}$

Alternative solution using L'Hospital's rule:
$\begin{align*}\lim_{x\to 0}\frac{\sin 4x}{\tan 5x} &=\lim_{x\to 0} \frac{4\cos 4x}{5\sec ^2 5x}\\
& =\frac {4}{5}\lim_{x\to 0} \left( \cos 4x \cdot \cos ^2 5x \right)\\
& =\frac {4}{5}\lim_{x\to 0} \left( \cos 4x \right) \cdot \lim_{x\to 0} \left( \cos ^2 5x \right)\\
& =\frac {4}{5}\cdot 1 \cdot 1=\frac{4}{5}\end{align*}$