<s>[INST] How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate: 
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}$$
But I don't know how to evaluate this?
Many thanks for any help.
 [/INST] Using the identity $sin \ x =3sin(\frac{x}{3})-4 sin^{3}(\frac{x}{3})$, our limit $l=lim \frac{sin \ x -x +\frac{x^{3}}{6}}{x^{3}}$ becomes, 
$l= \frac{1}{9} lim \frac{sin \ \frac{x}{3} -\frac{x}{3} +(\frac{x}{3})^{3}.\frac{1}{6}}{x^{3}}+ \frac{4}{27}- lim \  4 \frac{sin^{3}(\frac{x}{3})}{x^{3}}$, which means $l=\frac{1}{9} l$, so $l=0$.
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