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Question Number 152287 by john_santu last updated on 27/Aug/21

$$\underset{x\to 0}{\lim }{(1-\tan {}^2\mathrm{x})}^{\frac{6}{\sin {}^2\mathrm{x}}}=?$$

Answered by iloveisrael last updated on 27/Aug/21

$$\underset{x\to 0}{\lim }{(1-\tan {}^{2}x)}^{\frac{6}{\sin {}^{2}x}}=e^{\underset{x\to 0}{\lim }(1-\tan {}^{2}x-1).\frac{6}{\sin {}^{2}x}}$$$$=e^{\underset{x\to 0}{\lim }(-\tan {}^{2}x).\frac{6}{\sin {}^{2}x}}=e^{\underset{x\to 0}{\lim }(-6\cos {}^{2}x)}$$$$=e^{-6}=\frac{1}{e^6}.$$

Answered by john_santu last updated on 27/Aug/21

$$\underset{x\to 0}{\lim }{\left[{(1+(-\tan {}^2\mathrm{x}))}^{\frac{1}{-\tan {}^2\mathrm{x}}}\right]}^{-\frac{6\tan {}^2\mathrm{x}}{\sin {}^2\mathrm{x}}}$$$$={\mathrm{e}}^{\underset{x\to 0}{\lim }(-6\cos {}^2\mathrm{x})}={\mathrm{e}}^{-6}$$

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