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.

You can continue, using $\cos x = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = 1-2\sin^2(\frac{x}{2}) \iff \cos x -1 = -2\sin^2(\frac{x}{2})$ after your last step so that  $$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2} = \lim_{x\to0} \frac{1+2\Big(\frac{\cos x - 1}{x^2}\Big)}{6} = \lim_{x\to0} \frac{1+2\Big(\frac{-2\sin^2(\frac{x}{2})}{x^2}\Big)}{6} = \lim_{t\to0} \frac{1-\Big(\frac{\sin t}{t}\Big)^2}{6} = \lim_{t\to 0} \frac{1-1}{6} = \color{red}{\fbox{0}}$$
where $t=\frac{x}{2}$
