Answer
The solution is
$$\lim_{x\to 0}\frac{3\sin4x}{5x}=\frac{12}{5}.$$
Work Step by Step
We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to 0}\frac{3\sin4x}{5x}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to 0}\frac{(3\sin 4x)'}{(5x)'}=\lim_{x\to 0}\frac{3\cos 4x(4x)'}{5}=\lim_{x\to 0}\frac{12\cos4x}{5}=\left[\frac{12\cdot\cos 0}{5}\right]=\frac{12}{5}.$$
