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Tìm giới hạn của hàm số
[laTEX]\lim_{ x\to 0 }\dfrac{1 - cos3x.cos5x.cos7x}{sin^2x} [/laTEX]
Ta có
[laTEX]1 - cos3x.cos5x.cos7x = cos3x.cos5x - cos3x.cos5x.cos7x + 1 - cos3x.cos5x \\ \\ cos3x.cos5x(1 - cos7x) + cos3x- cos3x.cos5x + 1 - cos3x \\ \\ cos3x.cos5x(1 - cos7x) + cos3x(1-cos5x) + 1 - cos3x \\ \\ 2.sin^2(\frac{7x}{2})cos3x.cos5x+ 2cos3xsin^2(\frac{5x}{2}) + 2.sin^2(\frac{3x}{2}) \\ \\ \Rightarrow \lim_{ x\to 0 }\frac{2.sin^2(\frac{7x}{2})cos3x.cos5x+ 2cos3xsin^2(\frac{5x}{2}) + 2.sin^2(\frac{3x}{2})}{sin^2x} \\ \\ L_1+L_2+L_3 \\ \\ L_1 = \lim_{ x\to 0 }\frac{2.sin^2(\frac{7x}{2})cos3x.cos5x}{sin^2x} = \lim_{ x\to 0 }\frac{(\frac{7}{2})^22.x^2sin^2(\frac{7x}{2})cos3x.cos5x}{sin^2x(\frac{7x}{2})^2} \\ \\ L_1 = 2.\frac{49}{4} = \frac{49}{2}[/laTEX]
L_2 , L_3 làm tương tự
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