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Question Number 195401 by mnjuly1970 last updated on 01/Aug/23

$$$$$$$$$$\mathrm{calc}\underset{\underset{\Downarrow }{\Downarrow }}{\mathrm{u}late}$$$$$$$${\lim }_{\mathrm{x}\to \frac{\pi }{2}}{\left(\sin \left(\mathrm{x}\right)\right)}^{{\tan }^2\left(\mathrm{x}\right)}=?\overline{)\begin{array}{c}\\ \end{array}}$$$$$$$$$$

Answered by MM42 last updated on 01/Aug/23

$$e^{{lim}_{x\to \frac{\pi }{2}}(sinx-1){tan}^{2}x}=e^{{lim}_{x\to \frac{\pi }{2}}\frac{(sinx-1){sin}^{2}x}{{cos}^{2}x}}$$$$hop=e^{{lim}_{x\to \frac{\pi }{2}}\frac{\frac{1}{2}sin2xsinx+(sinx-1)sin2x}{-sin2x}}$$$$=e^{{lim}_{x\to \frac{\pi }{2}}\frac{\frac{1}{2}sinx+(sinx-1)}{-1}}=e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}✓$$$$$$$$$$

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