<s>[INST] Compute the limit $\displaystyle \lim_{x\rightarrow 0}\frac{n!.x^n-\sin (x).\sin (2x).\sin (3x).......\sin (nx)}{x^{n+2}}\;\;,$ 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 $n\in\mathbb{N}$
 [/INST] We know for small $y,$  $$\sin y=y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots=y\left(1-\frac{y^2}{3!}+\frac{y^4}{5!}-\cdots\right)$$ 
So, $$\prod_{1\le r\le n}\sin rx=\prod_{1\le r\le n}rx\left(1-\frac{(rx)^2}{3!}+\frac{(rx)^4}{5!}+\cdots\right)=n!x^n\prod_{1\le r\le n}\left(1-\frac{(rx)^2}{3!}+\frac{(rx)^4}{5!}+\cdots\right)$$
$$=n!x^n\prod_{1\le r\le n}\left( 1-\frac1{3!} r^2x^2 +O(x^4) \right)$$
So, $$n!x^n-\prod_{1\le r\le n}\sin rx=n!x^n \left(\frac{x^2}{3!}(1^2+2^2+\cdots+n^2) +O(x^4)\right)$$
So, $$\lim_{x\to0}\frac{n!x^n-\prod_{1\le r\le n}\sin rx}{x^{n+2}}=n!\frac{1^2+2^2+\cdots+n^2}{3!}=n!\frac{n(n+1)(2n+1)}{36}$$
</s>