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...

By Taylor series $$\cos x \sim_0 1 - \frac{x^2}{2}, \,\sin 2x \sim_0 2x\, \text{ and }\,\Big(1-\frac{x^2}{2}\Big)^3 \sim_0 1 - \frac{3x^2}{2}$$ so $$\dfrac{1-\cos^3 x}{x\sin2x} \sim_0 \frac{1 - \Big(1-\frac{x^2}{2}\Big)^3}{2x^2} \sim_0 \frac{ \frac{3x^2}{2} }{2x^2} = \frac{3}{4}.$$
