lim _{x rightarrow 0} frac{x tan 2 x-2 x tan | vedclass

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Question

(C) We have the expression $L = \lim _{x ightarrow 0} \frac{x 	an 2 x-2 x 	an x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}$.First,simplify th

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$\frac{16}{9}$

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(C) We have the expression $L = \lim _{x ightarrow 0} \frac{x 	an 2 x-2 x 	an x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}$.First,simplify the denominator term $(\operatorname{cosec} x-\cot x) = \frac{1-\cos x}{\sin x} = \frac{2 \sin^2(x/2)}{2 \sin(x/2) \cos(x/2)} = 	an(x/2)$.So,$(\operatorname{cosec} x-\cot x)^2 = 	an^2(x/2) \approx (x/2)^2 = \frac{x^2}{4}$ as $x ightarrow 0$.Also,$1-\cos 3x = 2 \sin^2(\frac{3x}{2}) \approx 2(\frac{3x}{2})^2 = \frac{9x^2}{2}$.The denominator becomes $\frac{9x^2}{2} \cdot \frac{x^2}{4} = \frac{9x^4}{8}$.Now,simplify the numerator: $x 	an 2x - 2x 	an x = x(2x + \frac{(2x)^3}{3} + \dots) - 2x(x + \frac{x^3}{3} + \dots) = 2x^2 + \frac{8x^4}{3} - 2x^2 - \frac{2x^4}{3} = \frac{6x^4}{3} = 2x^4$.Thus,$L = \lim _{x ightarrow 0} \frac{2x^4}{9x^4/8} = 2 \cdot \frac{8}{9} = \frac{16}{9}$.

$\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}=$

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