Question Number 213821 by hardmath last updated on 17/Nov/24
$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)^{\frac{\mathrm{sinx}}{\mathrm{x}\:−\:\mathrm{sinx}}} \:\:=\:\:? \\ $$
Answered by mehdee7396 last updated on 17/Nov/24
$${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{{sinx}}{{x}}−\mathrm{1}\right)\frac{{sinx}}{{x}−{sinx}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{{sinx}−{x}}{{x}}\right)\frac{{sinx}}{{x}−{sinx}}=−\mathrm{1} \\ $$$$\Rightarrow{answer}={e}^{−\mathrm{1}} \: \\ $$$$ \\ $$
Answered by lepuissantcedricjunior last updated on 20/Nov/24
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)^{\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}} =\mathrm{1}^{\boldsymbol{{FI}}} \\ $$$$=>\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)^{\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\boldsymbol{{sinx}}\:\boldsymbol{{ln}}\left(\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}} \\ $$$$\boldsymbol{{dl}}\:\boldsymbol{{sinx}}\:\grave {\boldsymbol{{a}}}\:\boldsymbol{{l}}'{o}\boldsymbol{{rdre}}\:\mathrm{3}\:\:\boldsymbol{{au}}\:\boldsymbol{{vois}}\:\boldsymbol{{de}}\:\mathrm{0} \\ $$$$\boldsymbol{{sinx}}=\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\frac{\boldsymbol{{x}}^{\boldsymbol{{k}}} }{\boldsymbol{{k}}!}×\boldsymbol{{f}}^{\boldsymbol{{k}}} \left(\mathrm{0}\right)=\boldsymbol{{x}}−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}+\boldsymbol{{x}}\left(\mathrm{0}\right)^{\mathrm{3}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\left(\boldsymbol{{x}}−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}\right)\boldsymbol{{ln}}\left(\mathrm{1}−\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{6}}\right)}{\boldsymbol{{x}}−\boldsymbol{{x}}+\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}}} \boldsymbol{{or}}\:\boldsymbol{{dl}}\:\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{t}}\right)\:\grave {\boldsymbol{{a}}}\:\boldsymbol{{l}}'\boldsymbol{{ordre}}\:\mathrm{3}\:\boldsymbol{{au}}\:\boldsymbol{{vois}}\:\boldsymbol{{de}}\:\mathrm{0}\:\boldsymbol{{donne}} \\ $$$$\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{t}}\right)=−\boldsymbol{{t}}+\frac{\boldsymbol{{t}}^{\mathrm{2}} }{\mathrm{2}}−\frac{\boldsymbol{{t}}^{\mathrm{3}} }{\mathrm{3}}+\boldsymbol{{t}}\left(\mathrm{0}\right)^{\mathrm{3}} \\ $$$$=>\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)^{\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\left(\boldsymbol{{x}}−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}\right)\left(−\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{6}}+\frac{\boldsymbol{{x}}^{\mathrm{4}} }{\mathrm{12}}\right)}{\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}}{\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}}} =\boldsymbol{{e}}^{−\mathrm{1}} \\ $$$$…………………………………………………. \\ $$$${prof}\:{cedric}\:{junior}….. \\ $$