<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] Otherwise using:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
$$\eqalign{
\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}
&=\lim_{x\to 0} \dfrac{(1-\cos x)(1+\cos x+\cos^2x)}{2x\sin x\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x\sin x\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{\frac{(1-\cos x)}x}{x\frac{\sin x}x\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x^2\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x^2\cos x}.\frac{1+\cos x}{1+\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{1}{\cos x(1+\cos x)}.\left(\frac{\sin x}{x}\right)^2\\
&=\frac32\times\frac1{1\times2}\times1^2=\frac34.}$$

Using Taylor series:
$$\cos x =1-x^2/2+O(x^4)$$
$$\sin x =x-x^3/6+O(x^5)$$
$$(1+x)^n=1+nx+O(x^2)$$
So,
$$\eqalign{
  \lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}
&=\lim_{x\to 0} \dfrac{1-(1-x^2/2+O(x^4))^3}{x(2x+O(x^3))}\\
&=\lim_{x\to 0} \dfrac{1-(1-3x^2/2+O(x^3))}{(2x^2+O(x^3))}\\
&=\lim_{x\to 0} \dfrac{3x^2/2+O(x^3)}{2x^2+O(x^3)}=\frac34.}$$ 

Using L-Hospital Rule:
$$\eqalign{\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}&
=\lim_{x\to 0} \dfrac{3\cos^2 x.\sin x}{\sin2x+2x\cos2x}\\
&=\lim_{x\to 0} \dfrac{3\sin x}{\sin2x+2x\cos2x}\\
&=\lim_{x\to 0} \dfrac{3\cos x}{2\cos2x+2\cos2x-4x\sin2x}\\
&=\frac{3\times1}{2\times1+2\times1-4\times0}=\frac34.}$$
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