Chủ đề này có 1 trả lời

[songviae]

Tính $\lim_{x\rightarrow 0}\frac{1-cosx.cos2x.cos3x...cosnx}{x^{2}}$

 

[trambau]

Tính $\lim_{x\rightarrow 0}\frac{1-cosx.cos2x.cos3x...cosnx}{x^{2}}$

Dùng quy tắc L'hospital  $\lim_{x\to0}\frac{\sin x}{x}=1\implies \lim_{x\to0}\frac{1-\cos nx}{x^2}=\lim_{x\to0}\left [\frac{2\sin^2\frac {nx}2}{\frac {n^2x^2}{2^2}}  \cdot\frac{n^2}{2^2}\right ]=\frac{n^2}{2}\left (\lim_{x\to0}\frac{\sin\frac {nx}2}{\frac {nx}2}  \right )^2=\frac{n^2}{2}$

$\Rightarrow \lim_{x\to 0}\frac{1-\cos x.\cos2x...\cos nx}{x^{2}}= \lim_{x\to0}\frac{1-\cos x+\cos x-\cos x\cos2x+\cos x\cos2x+\dots+\cos x.\cos2x...\cos (n-1)x-\cos x.\cos2x...\cos nx}{x^2}=\lim_{x\to0} \frac{1-\cos x+\cos x\left (1-\cos2x \right )+\dots+\cos x.\cos2x...\cos (n-1)x\left (1-\cos nx \right )}{x^2}=\lim_{x\to0}\frac{1-\cos x}{x^2}+\lim_{x\to0}\frac{1-\cos 2x}{x^2}+\cdots+\lim_{x\to0}\frac{1-\cos nx}{x^2}$ $ = \frac{1^2}{2}+\frac{2^2}{2}+\cdots+\frac{n^2}{2}=\frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{12}$