L-lim-x-pi-4-1-cos-2-x-2tanx-cos2x-

Question Number 120914 by SOMEDAVONG last updated on 04/Nov/20

L / . \lim_{x \to \frac{π}{4}} \frac{\frac{1}{\cos^{2} x} - 2 tanx}{\cos 2 x} = ?

Answered by bemath last updated on 04/Nov/20

\lim_{x \to π / 4} \frac{\sec^{2} x - 2 \tan x}{\cos 2 x}

set x = \frac{π}{4} + X

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

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

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

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

\lim_{X \to 0} \frac{1}{{(1 - \tan X)}^{2}} . \lim_{X \to 0} \frac{4 \tan^{2} X}{\sin 2 X} =

1 \times 0 = 0 .

Answered by Jamshidbek2311 last updated on 04/Nov/20

\lim_{x \to \frac{π}{4}} \frac{1 + {t a n}^{2} x - 2 t a n x}{c o s 2 x} = \frac{{(t a n x - 1)}^{2}}{c o s 2 x}

Missing \left or extra \right

Answered by MJS_new last updated on 04/Nov/20

let t = \tan \frac{x}{2}

\frac{\frac{1}{\cos^{2} x} - 2 \tan x}{\cos 2 x} = \frac{t^{2} + 1 - 2 t}{- \frac{t^{2} - 1}{t^{2} + 1}} = \frac{(t - 1) (t^{2} + 1)}{t + 1}

\lim_{t \to 1} - \frac{(t - 1) (t^{2} + 1)}{t + 1} = 0

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