Question Number 195401 by mnjuly1970 last updated on 01/Aug/23
$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{calc}\underset{\underset{\Downarrow} {\Downarrow}} {\mathrm{u}late} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\left(\:\:\mathrm{sin}\left(\mathrm{x}\:\right)\right)^{\:\mathrm{tan}^{\:\mathrm{2}} \left(\mathrm{x}\:\right)} \:\:\:\:=\:?\:\:\:\:\:\:\:\:\begin{array}{|c|c|}\\\\\hline\end{array} \\ $$$$ \\ $$$$ \\ $$
Answered by MM42 last updated on 01/Aug/23
$${e}^{{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\left({sinx}−\mathrm{1}\right){tan}^{\mathrm{2}} \:{x}} ={e}^{{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\frac{\left({sinx}−\mathrm{1}\right){sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}} \\ $$$${hop}\:={e}^{{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\frac{\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{xsinx}+\left({sinx}−\mathrm{1}\right){sin}\mathrm{2}{x}}{−{sin}\mathrm{2}{x}}} \\ $$$$={e}^{{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\frac{\frac{\mathrm{1}}{\mathrm{2}}{sinx}+\left({sinx}−\mathrm{1}\right)}{−\mathrm{1}}} ={e}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{{e}}}\:\checkmark \\ $$$$ \\ $$$$ \\ $$