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Question Number 114539 by naka3546 last updated on 19/Sep/20

$$\underset{x\to 0}{\lim }\left[\frac{\sinh x-\sin x}{x^3}\right]=?$$

Commented by bemath last updated on 20/Sep/20

$$\underset{x\to 0}{\lim }\frac{\frac{1}{2}[(1+x+\frac{x^2}{2}+\frac{x^3}{6})-(1-x+\frac{x^2}{2}-\frac{x^3}{6})]-(x-\frac{x^3}{6})}{x^3}$$$$\underset{x\to 0}{\lim }\frac{\frac{1}{2}(2x+\frac{x^3}{3})-(x-\frac{x^3}{6})}{x^3}$$$$\underset{x\to 0}{\lim }\frac{\frac{2}{6}{x}^3}{x^3}=\frac{1}{3}.$$

Answered by bobhans last updated on 19/Sep/20

$$\underset{x\to 0}{\lim }\frac{\frac{e^x-e^{-x}}{2}-\sin x}{x^3}=\underset{x\to 0}{\lim }\frac{\frac{e^x+e^{-x}}{2}-\cos x}{3x^2}$$$$\underset{x\to 0}{\lim }\frac{\frac{e^x-e^{-x}}{2}+\sin x}{6x}=\underset{x\to 0}{\lim }\frac{\frac{e^x+e^{-x}}{2}+\cos x}{6}$$$$=\frac{1}{3}$$

Answered by Dwaipayan Shikari last updated on 19/Sep/20

$$\underset{x\to 0}{\lim }\left(\frac{\frac{e^x-e^{-x}}{2}-x+\frac{x^3}{3}}{x^3}\right)=\underset{x\to 0}{\lim }\frac{e^x-e^{-x}-2x}{x^3}+\frac{1}{3}$$$$=\frac{e^x+e^{-x}-2}{3x^2}+\frac{1}{3}$$$$=\frac{e^x-e^{-x}}{6x}+\frac{1}{3}=\frac{e^x+e^{-x}}{6}+\frac{1}{3}=\frac{1}{3}$$

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