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Question Number 146043 by liberty last updated on 10/Jul/21

Answered by iloveisrael last updated on 10/Jul/21

$$\left(2\right)\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^{\sin \mathrm{x}}-1-\sin \mathrm{x}}{{\left({\tan }^{-1}\left(\sin \mathrm{x}\right)\right)}^2}$$$$\mathrm{let}{\tan }^{-1}\left(\sin \mathrm{x}\right)=\mathrm{u}\to \sin \mathrm{x}=\tan \mathrm{u}$$$$\underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{\tan \mathrm{u}}-1-\tan \mathrm{u}}{{\mathrm{u}}^2}=$$$$\underset{\mathrm{u}\to 0}{\lim }\frac{\sec {}^2\mathrm{u}{\mathrm{e}}^{\tan \mathrm{u}}-\sec {}^2\mathrm{u}}{2\mathrm{u}}=$$$$\underset{\mathrm{u}\to 0}{\lim }\frac{\sec {}^2\mathrm{u}}{2}.\underset{\mathrm{u}\to 0}{\lim }\frac{{\mathrm{e}}^{\tan \mathrm{u}}-1}{\mathrm{u}}=$$$$\frac{1}{2}.\underset{\mathrm{u}\to 0}{\lim }\frac{\sec {}^2\mathrm{u}{\mathrm{e}}^{\tan \mathrm{u}}}{1}=\frac{1}{2}\times 1=\frac{1}{2}$$

Answered by iloveisrael last updated on 10/Jul/21

$$\left(1\right)\underset{x\to \frac{3\pi }{4}}{\lim }\frac{1+\sqrt[3]{\tan \mathrm{x}}}{1-2\cos {}^2\mathrm{x}}$$$$=\underset{x\to \frac{3\pi }{4}}{\lim }\frac{1+\tan \mathrm{x}}{1-2\cos {}^2\mathrm{x}}.\underset{x\to \frac{3\pi }{4}}{\lim }\frac{1}{1+\sqrt[3]{\tan {}^2\mathrm{x}}-\sqrt[3]{\tan \mathrm{x}}}$$$$=\frac{1}{3}\times \underset{x\to \frac{3\pi }{4}}{\lim }\frac{\cos \mathrm{x}+\sin \mathrm{x}}{\cos \mathrm{x}(-\cos 2\mathrm{x})}$$$$=-\frac{1}{3}\times \underset{x\to \frac{3\pi }{4}}{\lim }\frac{1}{\cos \mathrm{x}(\cos \mathrm{x}-\sin \mathrm{x})}$$$$=-\frac{1}{3}\times \frac{1}{-\frac{1}{\sqrt{2}}(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})}$$$$=\frac{1}{3}\times \frac{1}{\frac{1}{\sqrt{2}}.\frac{2}{\sqrt{2}}}=\frac{1}{3}$$

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