1. ## Limit

Hi everybody,

i must calculate these following limits:

1)-$\displaystyle \lim_{x \to 0} \frac{1-coxcos2x...cosnx}{x^2}$

2)-$\displaystyle \lim_{x \to 0}\frac{1-cosxcos^2(2x)...cos^n(nx)}{x^2}$

Can you help me please?

And thank you anyway.

2. 1) $\displaystyle L=\lim_{x\to 0}\frac{1-\cos x\cos 2x\ldots\cos nx}{x^2}=$

(add and substract $\displaystyle \cos x$ at numerator)

$\displaystyle =\lim_{x\to 0}\frac{1-\cos x+\cos x-\cos x\cos 2x\ldots\cos nx}{x^2}=$

$\displaystyle =\lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{\cos x(1-\cos 2x\cos 3x\ldots\cos nx)}{x^2}=$

$\displaystyle =\lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{1-\cos 2x\cos 3x\ldots\cos nx}{x^2}$

Do the same thing to the second limit: add and substract $\displaystyle \cos 2x$, and so on. Then

$\displaystyle L=\sum_{k=1}^n\lim_{x\to 0}\frac{1-\cos kx}{x^2}=\sum_{k=1}^n\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{x^2}=$

$\displaystyle =2\sum_{k=1}^n\frac{k^2}{4}=\frac{1}{2}\sum_{k=1}^ n k^2=\frac{n(n+1)(2n+1)}{12}$

2) Using the same method we have

$\displaystyle L=\lim_{x\to 0}\frac{1-\cos x\cos^22x\ldots\cos^nnx}{x^2}=$

$\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{1-\cos^22x}{x^2}+\lim_{x\to 0}\frac{1-\cos^33x}{x^2}+\ldots+\lim_{x\to 0}\frac{1-\cos^nnx}{x^2}=$

$\displaystyle =\sum_{k=1}^n\lim_{x\to 0}\frac{1-\cos^kkx}{x^2}=$

$\displaystyle =\sum_{k=1}^n\lim_{x\to 0}\frac{(1-\cos kx)(1+\cos kx+\cos^2kx+\ldots+\cos^{k-1}kx)}{x^2}=$

$\displaystyle =\sum_{k=1}^n\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{x^2}\cdot\lim_{x\to 0}(1+\cos kx+\cos^2kx+\ldots\cos^{k-1}kx)=$

$\displaystyle =\frac{1}{2}\sum_{k=1}^nk^3=\frac{1}{2}\left(\frac {n(n+1)}{2}\right)^2=\frac{n^2(n+1)^2}{8}$

3. Originally Posted by red_dog
1) $\displaystyle L=\lim_{x\to 0}\frac{1-\cos x\cos 2x\ldots\cos nx}{x^2}=$

(add and substract $\displaystyle \cos x$ at numerator)

$\displaystyle =\lim_{x\to 0}\frac{1-\cos x+\cos x-\cos x\cos 2x\ldots\cos nx}{x^2}=$

$\displaystyle =\lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{\cos x(1-\cos 2x\cos 3x\ldots\cos nx)}{x^2}=$

$\displaystyle =\lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{1-\cos 2x\cos 3x\ldots\cos nx}{x^2}$

Do the same thing to the second limit: add and substract $\displaystyle \cos 2x$, and so on. Then

$\displaystyle L=\sum_{k=1}^n\lim_{x\to 0}\frac{1-\cos kx}{x^2}=\sum_{k=1}^n\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{x^2}=$

$\displaystyle =2\sum_{k=1}^n\frac{k^2}{4}=\frac{1}{2}\sum_{k=1}^ n k^2=\frac{n(n+1)(2n+1)}{12}$

2) Using the same method we have

$\displaystyle L=\lim_{x\to 0}\frac{1-\cos x\cos^22x\ldots\cos^nnx}{x^2}=$

$\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}+\lim_{x\to 0}\frac{1-\cos^22x}{x^2}+\lim_{x\to 0}\frac{1-\cos^33x}{x^2}+\ldots+\lim_{x\to 0}\frac{1-\cos^nnx}{x^2}=$

$\displaystyle =\sum_{k=1}^n\lim_{x\to 0}\frac{1-\cos^kkx}{x^2}=$

$\displaystyle =\sum_{k=1}^n\lim_{x\to 0}\frac{(1-\cos kx)(1+\cos kx+\cos^2kx+\ldots+\cos^{k-1}kx)}{x^2}=$

$\displaystyle =\sum_{k=1}^n\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{x^2}\cdot\lim_{x\to 0}(1+\cos kx+\cos^2kx+\ldots\cos^{k-1}kx)=$

$\displaystyle =\frac{1}{2}\sum_{k=1}^nk^3=\frac{1}{2}\left(\frac {n(n+1)}{2}\right)^2=\frac{n^2(n+1)^2}{8}$
Thank you very much.