lim-x-pi-4-1-tan-3-x-cos-2x-

Question Number 140574 by john_santu last updated on 09/May/21

\lim_{x \to \frac{π}{4}} \frac{1 - \tan^{3} x}{\cos 2 x} = ?

Answered by bramlexs22 last updated on 09/May/21

\lim_{x \to \frac{π}{4}} \frac{(1 + \tan^{2} x) (1 - \tan^{3} x)}{1 - \tan^{2} x}

= \lim_{x \to \frac{π}{4}} \frac{1 + \tan^{2} x}{1 + \tan x} . \lim_{x \to \frac{π}{4}} \frac{(1 - \tan x) (\tan^{2} x + \tan x + 1)}{1 - \tan x}

= \frac{2}{2} \times \frac{3}{1} = 3 .

Answered by john_santu last updated on 09/May/21

\lim_{x \to \frac{π}{4}} \frac{\cos^{3} x - \sin^{3} x}{\cos^{3} x (\cos^{2} x - \sin^{2} x)}

= \lim_{x \to \frac{π}{4}} \frac{\cos^{2} x + \cos x . \sin x + \sin^{2} x}{\cos^{3} x (\cos x + \sin x)} . \lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{\cos x - \sin x}

= \frac{\frac{3}{2}}{\frac{\sqrt{2}}{4} (\sqrt{2})} = 3 .

Answered by EDWIN88 last updated on 09/May/21

x = \frac{π}{4} + z \Rightarrow \lim_{z \to 0} \frac{1 - \tan^{3} (z + \frac{π}{4})}{\cos (2 z + \frac{π}{2})}

= \lim_{z \to 0} \frac{(\tan^{2} (z + \frac{π}{4}) + \tan (z + \frac{π}{4}) + 1) (1 - \tan (z + \frac{π}{4}))}{- \sin 2 z}

= 3 \times \lim_{z \to 0} \frac{\cos (z + \frac{π}{4}) - \sin (z + \frac{π}{4})}{- \cos (z + \frac{π}{4}) \sin 2 z}

= - \frac{6}{\sqrt{2}} \times \lim_{z \to 0} \frac{- \sin (z + \frac{π}{4}) - \cos (z + \frac{π}{4})}{2 \cos 2 z}

= - \frac{6}{\sqrt{2}} \times \frac{- \sqrt{2}}{2} = 3

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