Question

Sa se calculeze limita:

$\lim_{x \to \ 0} \frac{n-(cosx+cos^2x+.............+cos^nx)}{sin^2x}$

Answer 1

$\lim\limits_{x\to 0}\dfrac{n-\sum\limits_{k=1}^{n} \cos ^k x}{\sin^2 x} \overset{\frac{0}{0}(L'H)}{=}\lim\limits_{x\to 0}\dfrac{0 -\sum\limits_{k=1}^{n}\Big(-k\cos ^{k-1} x\cdot \sin x\Big)}{2\sin x\cos x} = \\\\ \\ = \lim\limits_{x\to 0} \dfrac{\sin x\sum\limits_{k=1}^n\Big(k\cos^{k-1}x\Big)}{2\sin x\cos x} = \lim\limits_{x\to 0}\dfrac{\sum\limits_{k=1}^n\Big(k\cos^{k-1}x\Big)}{2\cos x} =\\\\ \\ =\dfrac{1\cdot 1+2\cdot 1+3\cdot 1+...+n\cdot 1}{2} = \boxed{\dfrac{n(n+1)}{4}}$