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?

You can use the Mean Value Theorem: for $x\in\mathbb{R}^*$, the function $\sin$ is continuous on $[0,x]$ and differentiable on $(0,x)$, hence there exists $c\in(0,x)$ such that
$$\sin(x)=x\cos(c).$$
Now, in the particular case where $x\in(0,\pi)$, we'll have $0<c<x<\pi$, and since $\cos$ is decreasing on $[0,\pi]$ we'll have
$$\cos(c)>\cos(x),$$
hence (since $x>0$),
$$x\cos(c)>x\cos(x).$$
We conclude that
$$\forall x\in(0,\pi),\ \sin(x)>x\cos(x).$$
Since $(0,\pi)$ is a punctured right sided neighborhood of $0$, we're done.
