Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...

After two applications of L'Hôpital's rule: $$ \displaystyle \begin{aligned} \lim_{x \to 0} \left( \dfrac{1-\cos^3 (x)}{x\sin (2x)} \right) & \ = \ \lim_{x \to 0} \left( \dfrac{3\sin (x) \cos^2 (x)}{\sin (2x) + 2x\cos (2x)} \right) \\ & \ = \ \lim_{x \to 0} \left( \dfrac{3\cos^3 (x) - 6\cos (x) \sin (x)}{4\cos (2x) - 4x \sin (2x)} \right) = \ ? \end{aligned} $$
