calculate-lim-x-0-ln-x-sinx-x-2-

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

calculate \lim_{x \to 0} \frac{\ln (\frac{x}{sinx})}{x^{2}}

Answered by liberty last updated on 18/Mar/21

\lim_{x \to 0} \frac{\ln x - \ln \sin x}{x^{2}} =

\lim_{x \to 0} \frac{\frac{1}{x} - \frac{\cos x}{\sin x}}{2 x} = \lim_{x \to 0} \frac{\sin x - x \cos x}{2 x^{2} \sin x}

= \lim_{x \to 0} \frac{x - \frac{x^{3}}{6} - x (1 - \frac{x^{2}}{2})}{2 x^{3}}

= \lim_{x \to 0} \frac{\frac{2 x^{3}}{6}}{2 x^{3}} = \frac{1}{6}

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