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Question Number 205379 by cortano12 last updated on 19/Mar/24

Answered by MM42 last updated on 19/Mar/24

$$hop\to ={lim}_{x\to 0}\frac{-cosx\times sin\left(sinx\right)+sinx}{4x^3}$$$$hop\to =\frac{sinx\times sin\left(sinx\right)-{cos}^{2}x\times cos\left(sinx\right)+cosx}{12x^2}$$$$={lim}_{x\to 0}\frac{cosx(1-cosx)}{12x^2}$$$$={lim}_{x\to 0}\frac{\frac{1}{2}{x}^2}{12}=\frac{1}{24}✓$$$$$$

Commented by Frix last updated on 20/Mar/24

$$\mathrm{I}\mathrm{get}\frac{1}{6}$$

Answered by namphamduc last updated on 22/Mar/24

$$$$$$L=\underset{x\to 0}{\lim }\frac{\cos \left(\sin \left(x\right)\right)-\cos \left(x\right)}{x^4}=-2\underset{x\to 0}{\lim }\frac{\sin \left(\frac{\sin \left(x\right)+x}{2}\right)\sin \left(\frac{\sin \left(x\right)-x}{2}\right)}{x^4}$$$$=-2\underset{x\to 0}{\lim }\frac{\sin \left(\frac{\sin \left(x\right)+x}{2}\right)}{\frac{\sin \left(x\right)+x}{2}}\cdot \frac{\sin \left(\frac{\sin \left(x\right)-x}{2}\right)}{\frac{\sin \left(x\right)-x}{2}}.\frac{\sin \left(x\right)+x}{2x}.\frac{\sin \left(x\right)-x}{2x^3}$$$$\underset{x\to 0}{\lim }\frac{\sin \left(x\right)-x}{x^3}=-\frac{1}{6}\mathrm{is}\mathrm{well}-\mathrm{known}$$$$\Rightarrow L=-2.1.1.\frac{1+1}{2}(-\frac{1}{6})\frac{1}{2}=\frac{1}{6}$$

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