Elementary Question about limits Prove 
$$
\lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} = 0
$$
Obviously, Using l'Hôpitals rule, we can evaluate this limit. But, taking derivatives of such functions is such a mess. Anyone sees a trick to do this problem faster? any ideas?

$$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \dfrac{x - \sin(x)}{x} \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x}}$$
$$1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x} = 1 - \left(1 - \dfrac{x}2 +\dfrac{x^2}3 - \cdots \right) - \left( \dfrac{\dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \cdots}{x} \right) \\ = \left(\dfrac{x}2 - \dfrac{x^2}3 + \cdots \right) - \left(\dfrac{x}2 - \dfrac{x^3}{4!} + \cdots \right) = -\dfrac{x^2}3 + \mathcal{O}(x^4)$$
$$\dfrac{x - \sin(x)}{x} = \dfrac{x^2}{3!} + \mathcal{O}(x^4)$$
Hence,$$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \left(\dfrac{x^2}{3!} + \mathcal{O}(x^4) \right) \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{-\dfrac{x^2}3 + \mathcal{O}(x^4)}$$
Now you should be able to finish it off.
