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Question Number 107714 by bemath last updated on 12/Aug/20

$$\frac{✓\mathcal{B}e\mathcal{M}ath✓}{cooll}$$$$\underset{x\to 0}{\lim }{(1-\tan x)}^{\frac{4}{\sin {}^{2}x}}=?$$

Answered by bemath last updated on 12/Aug/20

Answered by Dwaipayan Shikari last updated on 12/Aug/20

$$\underset{x\to 0}{\lim }{(1+(-tanx))}^{-\frac{1}{tanx}.(-\frac{4}{{sin}^{2}x}.tanx)}$$$$=e^{-\frac{4}{sinxcosx}}=e^p=0(p\to -\mathrm{\infty })$$

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