Question Number 91954 by john santu last updated on 04/May/20
$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{sin}\:\mathrm{2x}\right)}{\mathrm{ln}\left(\mathrm{sin}\:\mathrm{4x}\right)} \\ $$
Commented by jagoll last updated on 04/May/20
$$\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$
Commented by jagoll last updated on 04/May/20
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Commented by abdomathmax last updated on 04/May/20
$${hospital}\:{theorem} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{{ln}\left({sin}\left(\mathrm{2}{x}\right)\right)}{{ln}\left({sin}\left(\mathrm{4}{x}\right)\right)}\:={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{\frac{\mathrm{2}{cos}\left(\mathrm{2}{x}\right)}{{sin}\left(\mathrm{2}{x}\right)}}{\frac{\mathrm{4}{cos}\left(\mathrm{4}{x}\right)}{{sin}\left(\mathrm{4}{x}\right)}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{\mathrm{2}{cos}\left(\mathrm{2}{x}\right)}{{sin}\left(\mathrm{2}{x}\right)}×\frac{{sin}\left(\mathrm{4}{x}\right)}{\mathrm{4}{cos}\left(\mathrm{4}{x}\right)} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{cos}\left(\mathrm{2}{x}\right)}{{cos}\left(\mathrm{4}{x}\right)}×\frac{{sin}\left(\mathrm{4}{x}\right)}{{sin}\left(\mathrm{2}{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\mathrm{2}\:=\mathrm{1} \\ $$
Commented by jagoll last updated on 04/May/20
yes ������