Limit of a trig function. (Without using L'Hopital) I'm having trouble figuring out what to do here, I'm supposed to find this limit:
$$\lim_{x\rightarrow0} \frac{x\cos(x)-\sin(x)}{x^3}$$
But I don't know where to start, any hint would be appreciated, thanks!

Since$$x\cos(x)-\sin(x)=x\left(1-\frac{x^2}2+\cdots\right)-\left(x-\frac{x^3}6+\cdots\right)=-\frac{x^3}3+\cdots,$$your limit is equal to $-\dfrac13$.
