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Question Number 194344 by mathlove last updated on 04/Jul/23

Answered by qaz last updated on 04/Jul/23

$$\sin {}^{2}x={(x-\frac{1}{6}{x}^3+...)}^2=x^2-\frac{1}{3}{x}^4+...$$$$e^x+e^{-x}-x^2-2=\frac{1}{24}{x}^4+\frac{1}{24}{(-x)}^4+...=\frac{1}{12}{x}^4+...$$$$\underset{x\to 0}{lim}\frac{e^x+e^{-x}-x^2-2}{\sin {}^{2}x-x^2}=\underset{x\to 0}{lim}\frac{\frac{1}{12}{x}^4}{-\frac{1}{3}{x}^4}=-\frac{1}{4}$$

Answered by anurup last updated on 04/Jul/23

$$\underset{x\to 0}{\lim }\frac{e^x-e^{-x}-2x}{2\sin x\cos x-2x}\left[\mathrm{by}{\mathrm{L}}^{\prime }{\mathrm{Hospital}}^{\prime }\mathrm{s}\mathrm{Rule}\right]$$$$=\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^x-e^{-x}-2x}{\sin 2x-2x}$$$$=\underset{x\to 0}{\lim }\frac{e^x+e^{-x}-2}{2\cos 2x-2}\left[\mathrm{by}{\mathrm{L}}^{\prime }{\mathrm{Hospital}}^{\prime }\mathrm{s}\mathrm{rule}\right]$$$$=\underset{x\to 0}{\lim }\frac{e^x-e^{-x}}{-4\sin 2x}\left[\mathrm{by}{\mathrm{L}}^{\prime }{\mathrm{Hospital}}^{\prime }\mathrm{s}\mathrm{rule}\right]$$$$=\underset{x\to 0}{\lim }\frac{e^x+e^{-x}}{-8\cos 2x}\left[\mathrm{by}{\mathrm{L}}^{\prime }{\mathrm{Hospital}}^{\prime }\mathrm{s}\mathrm{rule}\right]$$$$=-\frac{1}{4}$$$$$$

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