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Question Number 136029 by mathmax by abdo last updated on 18/Mar/21
calculate lim_(x→0)  ((ln((x/(sinx))))/x^2 )
$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{sinx}}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$
Answered by liberty last updated on 18/Mar/21
 lim_(x→0)  ((ln x−ln sin x)/x^2 ) =  lim_(x→0)  (((1/x)−((cos x)/(sin x)))/(2x)) = lim_(x→0)  ((sin x−xcos x)/(2x^2 sin x))  = lim_(x→0)  ((x−(x^3 /6)−x(1−(x^2 /2)))/(2x^3 ))  = lim_(x→0)  (((2x^3 )/6)/(2x^3 )) = (1/6)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:{x}−\mathrm{ln}\:\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} }\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{{x}}−\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}}{\mathrm{2}{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}−{x}\mathrm{cos}\:{x}}{\mathrm{2}{x}^{\mathrm{2}} \mathrm{sin}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−{x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\mathrm{2}{x}^{\mathrm{3}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{6}}}{\mathrm{2}{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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