<s>[INST] Limit without L'Hopital's rule: $\lim_{x\to0} \frac{1-\cos^3 x}{x\sin2x}$ How can I solve the following problem without the use of the L'Hopitals's rule? 
$$\lim_{x\to0} \frac{1-\cos^3(x)}{x\sin{(2x)}}$$
 [/INST] Hint:
We have that $1-\cos^3x = (1-\cos x)(1+\cos x + \cos^2 x)$ and $x\sin 2x = 2x\sin x \cos x$, so by limit arithmetic
$$\lim_{x \to 0} \frac{1-\cos^3x}{x\sin 2x} = \lim_{x\to0} \frac{(1-\cos x)(1+\cos x + \cos^2 x)}{2x\sin x \cos x} = \frac{3}{2}\lim_{x \to 0}\frac{1-\cos x}{x\sin x}$$
From here, see if you can get a numerator in terms of $\sin x$.
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