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Question Number 174841 by infinityaction last updated on 12/Aug/22

$$\underset{x\to 0}{\lim }\frac{x\sin \left(\sin x\right)-{\sin }^{2}x}{x^6}$$

Answered by Ar Brandon last updated on 12/Aug/22

$$\mathcal{A}=\underset{x\to 0}{\lim }\frac{x\sin \left(\sin x\right)-{\sin }^{2}x}{x^6}$$$$=\underset{x\to 0}{\lim }\frac{x(\sin x-\frac{{\sin }^{3}x}{6})-{(x-\frac{x^3}{6})}^2}{x^6}$$$$=\underset{x\to 0}{\lim }\frac{x(x-\frac{x^3}{6}-\frac{1}{6}{(x-\frac{x^3}{6})}^3)-(x^2-\frac{x^4}{3}+\frac{x^6}{36})}{x^6}$$$$=\underset{x\to 0}{\lim }\frac{1}{x^6}\{x^2-\frac{x^4}{6}-\frac{1}{6}(x^4-\frac{x^6}{2}+\frac{x^8}{12}-\frac{x^{10}}{216})-x^2+\frac{x^4}{3}-\frac{x^6}{36}\}$$$$=\underset{x\to 0}{\lim }\frac{1}{x^6}(\frac{x^6}{18}-\frac{x^8}{72}+\frac{x^{10}}{1296})=\underset{x\to 0}{\lim }(\frac{1}{18}-\frac{x^2}{72}+\frac{x^4}{1296})=\frac{1}{18}$$

Commented by infinityaction last updated on 12/Aug/22

$$thankyousir$$

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