How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$

Using L'Hospital twice, $$ \lim_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}=\lim_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}=\lim_{x\to 0} \frac{x - \sin x }{6x}=\frac{1}{6}\lim_{x\to 0} \left(1-\frac{\sin x}{x}\right)=0 $$

You can use l'Hospital as many times as needed as long as the indeterminate forms conditions are fulfilled. In this case, using Taylor series can be helpful, too:

$$\sin x = x - \frac{x^3}6 + \frac{x^5}{120} - \ldots = x - \frac{x^3}6 + \mathcal O(x^5)$$

$$\implies \frac{\sin x - x + \frac{x^3}6}{x^3} = \frac{\mathcal O(x^5)}{x^3} = \mathcal O(x^2) \xrightarrow[x \to 0]{} 0$$