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Question Number 106956 by bemath last updated on 08/Aug/20

Answered by 1549442205PVT last updated on 08/Aug/20

$$\frac{\cos 2\mathrm{x}(1+{\sin }^2\mathrm{x})+{\cos }^2\mathrm{x}-2}{{\mathrm{x}}^2}$$$$=\frac{\cos 2\mathrm{x}(2-{\cos }^2\mathrm{x})+{\cos }^2\mathrm{x}-2}{{\mathrm{x}}^2}$$$$=\frac{2(\cos 2\mathrm{x}-1)+{\cos }^2\mathrm{x}(\cos 2\mathrm{x}+1)}{{\mathrm{x}}^2}$$$$=\frac{-2.{2\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}.{2\cos }^2\mathrm{x}}{{\mathrm{x}}^2}$$$$=\frac{-{4\sin }^2\mathrm{x}+{2\cos }^4\mathrm{x}}{{\mathrm{x}}^2}$$$$\mathrm{Hence},\underset{\mathrm{x}\to 0}{\lim }\frac{\cos 2\mathrm{x}(1+{\sin }^2\mathrm{x})+{\cos }^2\mathrm{x}-2}{{\mathrm{x}}^2}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{-{4\sin }^2\mathrm{x}+{2\cos }^4\mathrm{x}}{{\mathrm{x}}^2}=+\mathrm{\infty }$$

Commented by bemath last updated on 10/Aug/20

$$wrong$$

Answered by bemath last updated on 08/Aug/20

Answered by Dwaipayan Shikari last updated on 08/Aug/20

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

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