lim-x-0-1-tan-2-x-6-sin-2-x-

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

\lim_{x \to 0} {(1 - \tan^{2} x)}^{\frac{6}{\sin^{2} x}} = ?

Answered by iloveisrael last updated on 27/Aug/21

\lim_{x \to 0} {(1 - \tan^{2} x)}^{\frac{6}{\sin^{2} x}} = e^{\lim_{x \to 0} (1 - \tan^{2} x - 1) . \frac{6}{\sin^{2} x}}

= e^{\lim_{x \to 0} (- \tan^{2} x) . \frac{6}{\sin^{2} x}} = e^{\lim_{x \to 0} (- 6 \cos^{2} x)}

= e^{- 6} = \frac{1}{e^{6}} .

Answered by john_santu last updated on 27/Aug/21

\lim_{x \to 0} {[{(1 + (- \tan^{2} x))}^{\frac{1}{- \tan^{2} x}}]}^{- \frac{6 \tan^{2} x}{\sin^{2} x}}

= e^{\lim_{x \to 0} (- 6 \cos^{2} x)} = e^{- 6}