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Question Number 106410 by bemath last updated on 05/Aug/20

$$\underset{x\to 0}{\lim }\frac{\cos 2\mathrm{x}.(1+\sin {}^2\mathrm{x})+\cos {}^2\mathrm{x}-2}{{\mathrm{x}}^2}$$

Answered by john santu last updated on 05/Aug/20

$$\underset{x\to 0}{\lim }\frac{\cos 2\mathrm{x}(2-\cos {}^2\mathrm{x})+\cos {}^2\mathrm{x}-2}{{\mathrm{x}}^2}=$$$$\underset{x\to 0}{\lim }\frac{(2\cos {}^2\mathrm{x}-1)(2-\cos {}^2\mathrm{x})-(2-\cos {}^2\mathrm{x})}{{\mathrm{x}}^2}=$$$$\underset{x\to 0}{\lim }\frac{(2-\cos {}^2\mathrm{x})\{2\cos {}^2\mathrm{x}-1-1\}}{{\mathrm{x}}^2}=$$$$\underset{x\to 0}{\lim }\frac{2(\cos {}^2\mathrm{x}-1)(2-\cos {}^2\mathrm{x})}{{\mathrm{x}}^2}=$$$$\underset{x\to 0}{\lim }\frac{2(-\sin {}^2\mathrm{x})}{{\mathrm{x}}^2}\times \underset{x\to 0}{\lim }(2-\cos {}^2\mathrm{x})=$$$$-2\times 1=-2$$

Answered by Dwaipayan Shikari last updated on 05/Aug/20

$$\underset{x\to 0}{\lim }\frac{cos2x(1+{sin}^{2}x)-(1+{sin}^{2}x)}{x^2}$$$$\underset{x\to 0}{\lim }\frac{(cos2x-1)(1+{sin}^{2}x)}{x^2}=-2\frac{{sin}^{2}x}{x^2}.(1+{sin}^{2}x)=-2$$

Answered by mathmax by abdo last updated on 05/Aug/20

$$\mathrm{let}\mathrm{use}\mathrm{hospital}\mathrm{thelrem}\mathrm{u}\left(\mathrm{x}\right)=\cos \left(2\mathrm{x}\right)(1+{\sin }^2\mathrm{x})+{\cos }^2\mathrm{x}-2$$$$\mathrm{and}\mathrm{v}\left(\mathrm{x}\right)={\mathrm{x}}^2$$$$\mathrm{u}\left(\mathrm{x}\right)=\cos \left(2\mathrm{x}\right)(2-{\cos }^2\mathrm{x})+{\cos }^2\mathrm{x}-2$$$$=\cos \left(2\mathrm{x}\right)\{2-\frac{1+\cos \left(2\mathrm{x}\right)}{2}\}+\frac{1+\cos \left(2\mathrm{x}\right)}{2}-2$$$$=\cos \left(2\mathrm{x}\right)\left\{\frac{3-\cos \left(2\mathrm{x}\right)}{2}\right\}+\frac{1}{2}\cos \left(2\mathrm{x}\right)-\frac{3}{2}$$$$=\frac{3}{2}\cos \left(2\mathrm{x}\right)-\frac{1}{2}\left(\frac{1+\cos \left(4\mathrm{x}\right)}{2}\right)+\frac{1}{2}\cos \left(2\mathrm{x}\right)-\frac{3}{2}$$$$=2\cos \left(2\mathrm{x}\right)-\frac{1}{4}\cos \left(4\mathrm{x}\right)-\frac{1}{4}-\frac{3}{2}\Rightarrow {\mathrm{u}}^{{}^{\prime }}\left(\mathrm{x}\right)=-4\sin \left(2\mathrm{x}\right)+\sin \left(4\mathrm{x}\right)$$$$\mathrm{and}{\mathrm{u}}^{\left(2\right)}\left(\mathrm{x}\right)=-8\cos \left(2\mathrm{x}\right)+4\cos \left(2\mathrm{x}\right)\Rightarrow {\lim }_{\mathrm{x}\to 0}{\mathrm{u}}^{\left(2\right)}\left(\mathrm{x}\right)=-4$$$${\mathrm{v}}^{{}^{\prime }}\left(\mathrm{x}\right)=2\mathrm{x}\mathrm{and}{\mathrm{v}}^{\left(2\right)}\left(\mathrm{x}\right)=2\Rightarrow {\lim }_{\mathrm{x}\to 0}{\mathrm{v}}^{\left(2\right)}\left(\mathrm{x}\right)=2\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\frac{\mathrm{u}\left(\mathrm{x}\right)}{\mathrm{v}\left(\mathrm{x}\right)}=-2$$

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