Question Number 152287 by john_santu last updated on 27/Aug/21
$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\right)^{\frac{\mathrm{6}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}} \:=? \\ $$
Answered by iloveisrael last updated on 27/Aug/21
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}\right)^{\frac{\mathrm{6}}{\mathrm{sin}\:^{\mathrm{2}} {x}}} \:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}−\mathrm{1}\right).\frac{\mathrm{6}}{\mathrm{sin}\:^{\mathrm{2}} {x}}} \\ $$$$\:={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\mathrm{tan}\:^{\mathrm{2}} {x}\right).\frac{\mathrm{6}}{\mathrm{sin}\:^{\mathrm{2}} {x}}} \:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(−\mathrm{6cos}\:^{\mathrm{2}} {x}\right)} \\ $$$$\:=\:{e}^{−\mathrm{6}} \:=\:\frac{\mathrm{1}}{{e}^{\mathrm{6}} }\:. \\ $$
Answered by john_santu last updated on 27/Aug/21
![lim_(x→0) [(1+(−tan^2 x))^(1/(−tan^2 x)) ]^(−((6tan^2 x)/(sin^2 x))) = e^(lim_(x→0) (−6cos^2 x)) =e^(−6)](https://www.tinkutara.com/question/Q152289.png)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left(\mathrm{1}+\left(−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\right)\right)^{\frac{\mathrm{1}}{−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}} \:\right]^{−\frac{\mathrm{6tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}} \\ $$$$=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\mathrm{6cos}\:^{\mathrm{2}} \mathrm{x}\right)} =\mathrm{e}^{−\mathrm{6}} \: \\ $$