limx→0x2(tan2x−2tanx)2(1−cos2x)4=

limx→0x2(tan2x−2tanx)2(1−cos2x)4=limx→0x22tanx1−tan2x−2tanx2(2sin2x)4=limx→0x24tan2x1−1+tan2x1−tan2x216sin8x=14Ltx→0x2tan6xsin8x(1−tan2x)2=14Ltx→0tan6xx6sin8xx81(1−0)2=14.1=14

limx→0x2(tan2x−2tanx)2(1−cos2x)4=

limx→0x2(tan2x−2tanx)2(1−cos2x)4=limx→0x22tanx1−tan2x−2tanx2(2sin2x)4=limx→0x24tan2x1−1+tan2x1−tan2x216sin8x=14Ltx→0x2tan6xsin8x(1−tan2x)2=14Ltx→0tan6xx6sin8xx81(1−0)2=14.1=14

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

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Detailed Solution

$\underset{x \to 0}{l i m} \frac{x^{2} {(\tan 2 x - 2 \tan x)}^{2}}{{(1 - \cos 2 x)}^{4}} = \underset{x \to 0}{l i m} \frac{x^{2} {(\frac{2 \tan x}{1 - \tan^{2} x} - 2 \tan x)}^{2}}{{(2 \sin^{2} x)}^{4}} =$ $\underset{x \to 0}{l i m} \frac{x^{2} 4 \tan^{2} x {(\frac{1 - 1 + \tan^{2} x}{1 - \tan^{2} x})}^{2}}{16 \sin^{8} x}$

$= \frac{1}{4} \underset{x \to 0}{L t} \frac{x^{2} \tan^{6} x}{\sin^{8} x {(1 - \tan^{2} x)}^{2}} = \frac{1}{4} (\underset{x \to 0}{L t} \frac{(\frac{\tan^{6} x}{x^{6}})}{(\frac{\sin^{8} x}{x^{8}})}) \frac{1}{{(1 - 0)}^{2}} = \frac{1}{4} .1 = \frac{1}{4}$