Giới hạn I = l i m x → 0 (1 − c o s x . c o s 2 x . … . c o s 2024 x) / x^2 bằng (1) ________.

Đáp án

Giới hạn \(I = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{1 - {\rm{cos}}x.{\rm{cos}}2x.\,\, \ldots \,\,.{\rm{cos}}2024x}}{{{x^2}}}\) bằng (1) 1382935950.

Giải thích

Ta có: \(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{1 - {\rm{cos}}ax}}{{{x^2}}} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{2{\rm{si}}{{\rm{n}}^2}\left( {\frac{{ax}}{2}} \right)}}{{{x^2}}} = \mathop {{\rm{lim}}}\limits_{x \to 0} \left[ {{{\left( {\frac{{{\rm{sin}}\frac{{ax}}{2}}}{{\frac{{ax}}{2}}}} \right)}^2}.\frac{{{a^2}}}{2}} \right] = \frac{{{a^2}}}{2}\).

Áp dụng ta được:

\(I = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{{{x^2}}} + \cos x.\frac{{1 - \cos 2x}}{{{x^2}}} + \cos x\cos 2x.\frac{{1 - \cos 3x}}{{{x^2}}} +  \ldots  + \cos x.\,\,\, \ldots \,\,\,\left. {.\cos 2023x.\frac{{1 - \cos 2024x}}{{{x^2}}}} \right)} \right.\)

\( = \frac{{{1^2}}}{2} + \frac{{{2^2}}}{2} +  \ldots  + \frac{{{{2024}^2}}}{2} = \frac{{2024.\left( {2024 + 1} \right).\left( {2.2024 + 1} \right)}}{{12}} = 1382935950\)

(với \(n \in \mathbb{N},{1^2} + {2^2} +  \ldots  + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\))