Calculus: Finding limit $$\lim_{x\to0}\frac{\displaystyle\\x^3}{\sin^3x}.$$
Do I use L'Hopital's rule for this? But I can't seem to find the answer.

Hint: Since $x \mapsto x^3$ is continuous and $x \mapsto 1/x$ is too, we have $$\lim_{x \to 0 }\frac{x^3}{\sin^3x} = \left( \frac{1}{\color{red}{\lim_{x \to 0}\frac{\sin x}{x}}}\right)^3.$$That limit in red you absolutely must know.
