Dear Student, limx-0xtan2x - 2x tanx(1-cos2x)2=limx-0xsin2xcos2x - 2x sinxcosx(1-cos2x)2=limx-0x2sinxcos2x-2sinx cos2xcos2x cosx(2sin2x)2 =limx-0x2sinxcos2x-2sinx (cos2x-sin2x)cos2x cosx(2sin2x)2 =limx-0x2sinxcos2x-2sinx cos2x+2sin3x)cos2x cosx(2sin2x)2 =limx-0x2sin3xcos2x cosx(2sin2x)2=limx-0x2sin3xcos2x cosx4sin4x=limx-01cos2x cosx 2sinxx =11 1 2 1=12 option C)1/2 is the correct answer

Please solve 7th one

Dear Student,

\lim_{x - 0} \frac{xtan 2 x - 2 x tanx}{(1 - \cos 2 x)^{2}} = \lim_{x - 0} \frac{x \frac{\sin 2 x}{\cos 2 x} - 2 x \frac{sinx}{cosx}}{(1 - \cos 2 x)^{2}} = \lim_{x - 0} \frac{x \frac{(2 {sinxcos}^{2} x - 2 sinx . \cos 2 x)}{\cos 2 x . cosx}}{(2 \sin^{2} x)^{2}} = \lim_{x - 0} \frac{x \frac{(2 {sinxcos}^{2} x - 2 sinx . (\cos^{2} x - \sin^{2} x))}{\cos 2 x . cosx}}{(2 \sin^{2} x)^{2}} = \lim_{x - 0} \frac{x \frac{(2 {sinxcos}^{2} x - 2 sinx . \cos^{2} x + 2 \sin^{3} x))}{\cos 2 x . cosx}}{(2 \sin^{2} x)^{2}} = \lim_{x - 0} \frac{x \frac{2 \sin^{3} x}{\cos 2 x . cosx}}{(2 \sin^{2} x)^{2}} = \lim_{x - 0} \frac{x \frac{2 \sin^{3} x}{\cos 2 x . cosx}}{4 \sin^{4} x} = \lim_{x - 0} \frac{1}{\cos 2 x . cosx . 2 \frac{sinx}{x}} = \frac{1}{1 . 1 . 2 . 1} = \frac{1}{2}

option C)1/2 is the correct answer.

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