<s>[INST] 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...
 [/INST] $$\lim_{x \to 0}\frac{1-\cos^3x}{2x^2} \cdot \lim_{x \to 0}\frac{2x}{\sin(2x)}$$
The right limit is equal to 1.
For the left limit, you may use the rule of de l'Hôpital:
$$\eqalign{
\lim_{x \to 0}\frac{1-\cos^3x}{2x^2}&=\lim_{x \to 0}\frac{3\cos^2 x \sin x}{4x}\\
&=\lim_{x \to 0}\frac{-6\cos x \sin^2 x + 3 \cos^3 x}{4}\\
&=\frac{-6\cdot 0 + 3 \cdot 1}{4}\\
&=\frac{3}{4}.
}$$
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