Limx→0xtan2x-2xtanx(1-cos2x)2=

Ltx→0xtan2x-2xtanx(1-cos2x)2 =Ltx→0x 2Tan x1-Tan2x-2x Tan x(1-cos 2x)2 =Ltx→02x Tan x(1-1+Tan2x)(1-cos 2x)2(1-cos 2x)2 =Ltx→02x Tan3x(1-Tan2x)(1-cos 2x)2 =Ltx→02x Tan3xx4(1-Tan2x)1-cos 2xx22 =22222=2(2)2=12

Limx→0xtan2x-2xtanx(1-cos2x)2=

Ltx→0xtan2x-2xtanx(1-cos2x)2 =Ltx→0x 2Tan x1-Tan2x-2x Tan x(1-cos 2x)2 =Ltx→02x Tan x(1-1+Tan2x)(1-cos 2x)2(1-cos 2x)2 =Ltx→02x Tan3x(1-Tan2x)(1-cos 2x)2 =Ltx→02x Tan3xx4(1-Tan2x)1-cos 2xx22 =22222=2(2)2=12

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$\underset{x \to 0}{Lim} \frac{xtan 2 x - 2 xtanx}{{(1 - \cos 2 x)}^{2}} =$

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$\frac{- 1}{2}$

-2

$\frac{1}{2}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} \underset{x \to 0}{Lt} \frac{xtan 2 x - 2 xtanx}{{(1 - \cos 2 x)}^{2}} \\ = & \underset{x \to 0}{Lt} \frac{\frac{x 2 Tan x}{1 - {Tan}^{2} x} - 2 x Tan x}{{(1 - \cos 2 x)}^{2}} \\ = & \underset{x \to 0}{Lt} \frac{\frac{2 x Tan x (1 - 1 + {Tan}^{2} x)}{{(1 - \cos 2 x)}^{2}}}{{(1 - \cos 2 x)}^{2}} \\ = & \underset{x \to 0}{Lt} \frac{2 x {Tan}^{3} x}{(1 - {Tan}^{2} x) {(1 - \cos 2 x)}^{2}} \\ = & \underset{x \to 0}{Lt} \frac{\frac{2 x {Tan}^{3} x}{x^{4}}}{(1 - {Tan}^{2} x) {(\frac{1 - \cos 2 x}{x^{2}})}^{2}} \\ = & \frac{2}{{(\frac{2^{2}}{2})}^{2}} = \frac{2}{{(2)}^{2}} = \frac{1}{2} \end{array}$