Without L′Hopital lim_(x→(π/4)) ((((1/(cos^2 x)) −2tan x ))/(cos 2x)) ?


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Question Number 114754 by bemath last updated on 21/Sep/20

$W i t h o u t L^{'} H o p i t a l$ $\lim_{x \to \frac{π}{4}} \frac{(\frac{1}{\cos^{2} x} - 2 \tan x)}{\cos 2 x} ?$
Commented by bemath last updated on 21/Sep/20

$g a v e k u d o s a l l s i r$
	Answered by john santu last updated on 21/Sep/20

	$l e t x = \frac{π}{4} + h, h \to 0$ $\lim_{h \to 0} \frac{\sec^{2} (\frac{π}{4} + h) - 2 \tan (\frac{π}{4} + h)}{\cos (\frac{π}{2} + 2 h)} =$ $\lim_{h \to 0} \frac{\tan^{2} (\frac{π}{4} + h) - 2 \tan (\frac{π}{4} + h) + 1}{- \sin 2 h} =$ $\lim_{h \to 0} \frac{{(1 - \tan (\frac{π}{4} + h))}^{2}}{- \sin 2 h} =$ $\lim_{h \to 0} \frac{{[1 - (\frac{1 + \tan h}{1 - \tan h})]}^{2}}{- \sin 2 h} =$ $\lim_{h \to 0} \frac{{(- 2 \tan h)}^{2}}{- \sin 2 h} = \lim_{h \to 0} \frac{4 \tan^{2} h}{- \sin 2 h} = 0$
	Answered by Olaf last updated on 21/Sep/20

	$\lim_{x \to \frac{π}{4}} \frac{1 + \tan^{2} x - 2 \tan x}{\cos 2 x}$ $\lim_{x \to \frac{π}{4}} \frac{{(1 - \tan^{2} x)}^{2}}{\cos 2 x}$ $u = \frac{π}{4} - x$ $\lim_{u \to 0} \frac{{[1 - \tan {(\frac{π}{4} - u)}^{2}]}^{2}}{\cos (\frac{π}{2} - 2 u)}$ $\lim_{u \to 0} \frac{{[1 - {(\frac{\tan \frac{π}{4} - \tan u}{1 + \tan \frac{π}{4} \tan u})}^{2}]}^{2}}{\cos (\frac{π}{2} - 2 u)}$ $\lim_{u \to 0} \frac{{[1 - {(\frac{1 - \tan u}{1 + \tan u})}^{2}]}^{2}}{\sin 2 u}$ $\lim_{u \to 0} \frac{{[1 - {(\frac{1 - \tan u}{1 + \tan u})}^{2}]}^{2}}{\sin 2 u}$ $\lim_{u \to 0} \frac{{[\frac{{4 \tan}^{2} u}{{(1 + \tan u)}^{2}}]}^{2}}{\sin 2 u}$ $\lim_{u \to 0} \frac{{[\frac{4 u^{2}}{{(1 + u)}^{2}}]}^{2}}{2 u} = 0$
	Answered by bobhans last updated on 21/Sep/20

	$i n o t h e r w a y$ $\lim_{x \to \frac{π}{4}} \frac{{(\tan x - 1)}^{2}}{\cos^{2} x - \sin^{2} x} = \lim_{x \to \frac{π}{4}} \frac{{(\cos x - \sin x)}^{2}}{\cos^{2} x (\cos x - \sin x) (\cos x + \sin x)}$ $= \lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{\cos^{2} x (\cos x + \sin x)} = 0$
	Answered by MJS_new last updated on 21/Sep/20

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