\(\bf \lim\limits_{x\to 0}\ \cfrac{sin^3(x)}{5x^3}-4
\\ \quad \\\\ \cfrac{sin^3(x)}{5x^3}-4\implies \cfrac{[sin(x)]^3}{5x^3}-4
\\ \quad \\
\cfrac{[sin(x)]^3}{x^3}\cdot \cfrac{1}{5}-4
\\ \quad \\
thus
\\ \quad \\
\lim\limits_{x\to 0}\cfrac{[sin(x)]^3}{x^3}\cdot \lim\limits_{x\to 0}\cfrac{1}{5}-4\qquad \boxed{recall \qquad \lim\limits_{x\to 0} \cfrac{sin(x)}{x}\implies 1}
\\ \quad \\
1\cdot \cfrac{1}{5}-4\)