Lim x- 0 1 -cos 2x22x tan x - x tan 2x is

The correct option is B 2 lim_x→ 0(2sin^2x)^22xsin xcos x xsin 2xcos 2x=4sin^4x×cos xcos 2x2xcos 2xsin x xsin 2xcos x =4sin^4xxsin x(2cos 2x 2cos^2x) ...

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Byju's Answer

Standard XII

Mathematics

Limit

lim x→ 0 1 ...

Question

$\begin{matrix} l i m x \to 0 \end{matrix} \frac{(1 - c o s 2 x)^{2}}{2 x t a n x - x t a n 2 x}$ is

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- \frac{1}{2}

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\frac{1}{2}

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-2

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Solution

The correct option is B -2
$lim x \to 0 \frac{(2 {sin}^{2} x)^{2}}{\frac{2 x sin x}{cos x} - \frac{x sin 2 x}{cos 2 x}} = \frac{4 {sin}^{4} x \times cos x cos 2 x}{2 x cos 2 x sin x - x sin 2 x cos x}$

$= \frac{4 {sin}^{4} x}{x sin x (2 cos 2 x - 2 {cos}^{2} x)} = \frac{4 {sin}^{3} x}{(2 {cos}^{2} x - 2) x} = \frac{2 {sin}^{3} x}{- (1 - {cos}^{2} x) x}$

$= \frac{2 {sin}^{3} x}{- x {sin}^{2} x} = \frac{- 2 sin x}{x}$
$= - 2$

$\Rightarrow O p t i o n D$ .

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lim x \to 0 \frac{(1 - cos 2 x)^{2}}{2 x tan x - x tan 2 x}

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{lim}_{x \to 0} \frac{(1 - cos 2 x)^{2}}{2 x tan x - x tan 2 x}

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lim x \to 0 \frac{(1 - cos 2 x)^{2}}{2 x tan x - x tan 2 x}

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

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limx→0(1−cos2x)22xtanx−xtan2x is

The correct option is B -2limx→0(2sin2x)22xsinxcosx−xsin2xcos2x=4sin4x×cosxcos2x2xcos2xsinx−xsin2xcosx=4sin4xxsinx(2cos2x−2cos2x)=4sin3x(2cos2x−2)x=2sin3x−(1−cos2x)x=2sin3x−xsin2x=−2sinxx=−2⇒OptionD.

$\begin{matrix} l i m x \to 0 \end{matrix} \frac{(1 - c o s 2 x)^{2}}{2 x t a n x - x t a n 2 x}$ is