On proving that $\sin x > x \cos x$ in a right neighbourhood of zero. I was recently asked to prove rigorously that there exists a $\delta$ s.t. $\sin x > x \cos x$ in $]0,\delta[$ .
The best I could come up with was that Taylor expanding $\sin x - x  \cos x $ I obtain $\frac{ x^3}{3} + o(x^3)$ and noticing that the dominant term goes to zero for positive values I know that there must exist such a $\delta$.
But I was told this is not rigorous enough, how could I formalize it better? Any other paths I could take?

$${\sin x\over x}=\int_0^1\cos(t\,x)\>dt>\cos x\qquad(0<|x|\leq\pi)\ .$$
