Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n\in\mathbb{N}$

share|cite|improve this question
(+1) Cool question! – L. F. Jan 31 '13 at 4:34
up vote 4 down vote accepted

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, $$\lim_{x\to0}\frac{n!x^n-\prod_{1\le r\le n}\sin rx}{x^{n+2}}=n!\frac{1^2+2^2+\cdots+n^2}{3!}=n!\frac{n(n+1)(2n+1)}{36}$$

share|cite|improve this answer
Thanks lab bhattacharjee – juantheron Jan 31 '13 at 5:40
@juantheron, my pleasure. Hope, I could make the idea clear. – lab bhattacharjee Jan 31 '13 at 5:41

Since $\sin x = x - x^3/6 +O(x^5)$ as $x\to 0$, we get $$\begin{array} . & &\frac{n!x^n-\sin (x)\sin (2x)\sin (3x)\cdots\sin (nx)}{x^{n+2}} \\&=&\frac{n!x^n - (x-x^3/6+O(x^5))\cdots(nx-(nx)^3/6+O(x^5))}{x^{n+2}} \\&=& \frac{\frac{1}{6}x^{n+2}n! (1^2+2^2+\cdots+n^2)+O(x^{n+4})}{x^{n+2}} \end{array}$$ as $x\to0$. So desired limit is $\frac{1}{36}n!\cdot n(n+1)(2n+1)$.

share|cite|improve this answer
Thanks ......tetori – juantheron Jan 31 '13 at 5:41

$$\dfrac{\sin(kx)}{kx} = \left(1- \dfrac{k^2x^2}{3!} + \mathcal{O}(x^4)\right)$$ 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 $$\lim_{x \to 0} \dfrac{n!x^n - \displaystyle \prod_{k=1}^n \sin(kx)}{x^{n+2}} = n!\left(\lim_{x \to 0} \dfrac{1 - \displaystyle \prod_{k=1}^n \dfrac{\sin(kx)}{kx}}{x^{2}} \right) = \dfrac{n(2n+1)(n+1)!}{36}$$

share|cite|improve this answer
@labbhattacharjee I group $n!(n+1)$ as $(n+1)!$. – user17762 Jan 31 '13 at 4:27
sorry I missed that. – lab bhattacharjee Jan 31 '13 at 4:29
Thanks.... Marvis – juantheron Jan 31 '13 at 5:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.