Calculus

Using Taylor series expansion of $\sin(u) = u - \frac{u^3}{6} + O(u^5)$ with $u = 3x$: $\sin(3x) = 3x - \frac{(3x)^3}{6} + O(x^5) = 3x - \frac{27x^3}{6} + O(x^5) = 3x - \frac{9}{2}x^3 + O(x^5)$ Subtracting $3x$: $\sin(3x) - 3x = -\frac{9}{2}x^3 + O(x^5)$ Dividing by $x^3$ and taking limit $x \to 0$: $\lim_{x \to 0} \frac{-\frac{9}{2}x^3 + O(x^5)}{x^3} = -\frac{9}{2} = -4.5$