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Question Number 136029 by mathmax by abdo last updated on 18/Mar/21

$$\mathrm{calculate}{\lim }_{\mathrm{x}\to 0}\frac{\ln \left(\frac{\mathrm{x}}{\mathrm{sinx}}\right)}{{\mathrm{x}}^2}$$

Answered by liberty last updated on 18/Mar/21

$$\underset{x\to 0}{\lim }\frac{\ln x-\ln \sin x}{x^2}=$$$$\underset{x\to 0}{\lim }\frac{\frac{1}{x}-\frac{\cos x}{\sin x}}{2x}=\underset{x\to 0}{\lim }\frac{\sin x-x\cos x}{2x^2\sin x}$$$$=\underset{x\to 0}{\lim }\frac{x-\frac{x^3}{6}-x(1-\frac{x^2}{2})}{2x^3}$$$$=\underset{x\to 0}{\lim }\frac{\frac{2x^3}{6}}{2x^3}=\frac{1}{6}$$

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