Answer
$$\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x^{3}-4 x}=-\frac{3}{4}$$
Work Step by Step
Given $$\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x^{3}-4 x}$$
So,
\begin{align}
\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x^{3}-4 x}&=\lim _{x \rightarrow 0}\left(\frac{\sin 3 x}{3 x} \cdot \frac{3}{5 x^{2}-4}\right)\\
&=\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \cdot \lim _{x \rightarrow 0} \frac{3}{5 x^{2}-4}\\
&=1 \cdot\left(\frac{3}{-4}\right)=-\frac{3}{4}
\end{align}
