Question

Sa se calculeze limita:
$\lim_{x \to \0} \frac{1}{sin^x(nx)} *$ ∑ $(-1)^{k+1} * cos(kx)$
suma este de la k=1 la 2n

Answer 1

$\lim\limits_{x\to 0}\dfrac{\sum\limits_{k=1}^{2n}\Big[(-1)^{k+1}\cdot \cos(kx)\Big]}{\sin^x (nx)} =\\ \\\\=\lim\limits_{x\to 0}\dfrac{\sum\limits_{k=1}^{n}\Bigg[\cos\Big((2k-1)x\Big)-\cos\Big((2k)x\Big)\Bigg]}{\sin^x (nx)} =\\ \\= \lim\limits_{x\to 0}\left[\Bigg(\dfrac{nx}{\sin(nx)}\Bigg)^x\cdot \dfrac{\sum\limits_{k=1}^{n}\Bigg[\cos\Big((2k-1)x\Big)-\cos\Big((2k)x\Big)\Bigg]}{(nx)^x}\right] =$

$=\lim\limits_{x\to 0}\dfrac{ \sum\limits_{k=1}^{n}\Bigg[\cos\Big((2k-1)x\Big)-\cos\Big((2k)x\Big)\Bigg]}{(nx)^x} = \\ \\ = \lim\limits_{x\to 0}\dfrac{ \sum\limits_{k=1}^{n}\Bigg[\cos\Big((2k-1)x\Big)-\cos\Big((2k)x\Big)\Bigg]}{n^x\cdot x^x} = \\ \\ =\dfrac{\lim\limits_{x\to 0}\left(\sum\limits_{k=1}^{n}\Bigg[\cos\Big((2k-1)x\Big)-\cos\Big((2k)x\Big)\Bigg]\right)}{\lim\limits_{x\to 0}(n^x)\cdot\lim\limits_{x\to 0}(x^x)}$

$= \dfrac{\lim\limits_{x\to 0}\left(\sum\limits_{k=1}^{n}\Bigg[\cos\Big((2k-1)x\Big)-\cos\Big((2k)x\Big)\Bigg]\right)}{1\cdot 1}=\\ \\ = (1-1)+(1-1)+(1-1)+\underbrace{...}\limits_{de\,\,n\,\,ori}+(1-1) = \\ \\ = 0$