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Question Number 91419 by john santu last updated on 30/Apr/20

$$\underset{x\to 0}{\lim }\frac{1}{\sin {}^{4}x}(\sin \left(\frac{x}{1+x}\right)-\frac{\sin x}{1+\sin x})$$

Commented by john santu last updated on 01/May/20

$$\underset{x\to 0}{\lim }\frac{1}{\sin {}^{4}x}\left(\frac{(1+\sin x)\sin \left(\frac{x}{1+x}\right)-\sin x}{1+\sin x}\right)=$$$$\underset{x\to 0}{\lim }\frac{\sin \left(\frac{x}{1+x}\right)+\sin x\sin \left(\frac{x}{1+x}\right)-\sin x}{\sin {}^{4}x}=$$$$\underset{x\to 0}{\lim }\frac{\sin x\sin \left(\frac{x}{1+x}\right)+(2\cos \left(\frac{x(x+2)}{2(x+1)}\right)\sin \left(\frac{-x^2}{2(x+1)}\right)}{\sin {}^{4}x}=$$$$\underset{x\to 0}{\lim }\frac{\frac{x^2}{1+x}-\frac{x^2}{1+x}}{x^4}=0$$

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