TITLE: Compute the limit $\displaystyle \lim_{x\rightarrow 0}\frac{n!.x^n-\sin (x).\sin (2x).\sin (3x).......\sin (nx)}{x^{n+2}}\;\;,$
QUESTION [7 upvotes]: How can i calculate the Given limit
$\displaystyle \lim_{x\rightarrow 0}\frac{n!x^n-\sin (x)\sin (2x)\sin (3x)\dots\sin (nx)}{x^{n+2}}\;\;,$ where $n\in\mathbb{N}$

REPLY [2 votes]: $$\dfrac{\sin(kx)}{kx} = \left(1- \dfrac{k^2x^2}{3!} + \mathcal{O}(x^4)\right)$$
Hence,
$$\prod_{k=1}^n \dfrac{\sin(kx)}{kx} = \prod_{k=1}^n\left(1- \dfrac{k^2x^2}{3!} + \mathcal{O}(x^4)\right) = 1 - \dfrac{\displaystyle \sum_{k=1}^n k^2}6x^2 + \mathcal{O}(x^4)\\ = 1 - \dfrac{n(n+1)(2n+1)}{36}x^2 + \mathcal{O}(x^4)$$
Hence, the limit you have is
$$\lim_{x \to 0} \dfrac{n!x^n - \displaystyle \prod_{k=1}^n \sin(kx)}{x^{n+2}} = n!\left(\lim_{x \to 0} \dfrac{1 - \displaystyle \prod_{k=1}^n \dfrac{\sin(kx)}{kx}}{x^{2}} \right) = \dfrac{n(2n+1)(n+1)!}{36}$$