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Question Number 161747 by cortano last updated on 22/Dec/21

$$\underset{x\to 0}{\lim }\frac{\cos {}^3\left(2x\right)-\cos \left(x\right)}{\cos {}^2\left(4x\right)-\cos \left(2x\right)}=?$$

Answered by Ar Brandon last updated on 22/Dec/21

$$\mathcal{L}=\underset{x\to 0}{\lim }\frac{{\cos }^3\left(2x\right)-\cos \left(x\right)}{{\cos }^2\left(4x\right)-\cos \left(2x\right)}=\underset{x\to 0}{\lim }\frac{{(1-\frac{4x^2}{2})}^3-(1-\frac{x^2}{2})}{{(1-\frac{16x^2}{2})}^2-(1-\frac{4x^2}{2})}$$$$=\underset{x\to 0}{\lim }\frac{(1-3\frac{4x^2}{2}+3\frac{16x^4}{4}-\frac{64x^6}{8})-(1-\frac{x^2}{2})}{(1-2\frac{16x^2}{2}+\frac{256x^4}{4})-(1-\frac{4x^2}{2})}$$$$=\underset{x\to 0}{\lim }\frac{-\frac{11x^2}{2}+12x^4-8x^6}{-14x^2+64x^4}=\underset{x\to 0}{\lim }\frac{-\frac{11}{2}+12x^2-8x^4}{-14+64x^2}=\frac{11}{28}$$

Answered by blackmamba last updated on 22/Dec/21

$$\underset{x\to 0}{\lim }\frac{\cos {}^3\left(2x\right)-1+1-\cos \left(x\right)}{\cos {}^2\left(4x\right)-1+1-\cos \left(2x\right)}$$$$=\underset{x\to 0}{\lim }\frac{3(\cos 2x-1)+2\sin {}^2\left(\frac{x}{2}\right)}{2(\cos \left(4x\right)-1)+2\sin {}^2\left(x\right)}$$$$=\frac{3(-2)+\frac{1}{2}}{2(-2)\left(4\right)+2}=\frac{-\frac{11}{2}}{-14}=\frac{11}{28}$$

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