lim _{x rightarrow 0} frac{(1-cos 2 x)}{x tan | vedclass

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(B) Given,$\lim _{x ightarrow 0} \frac{1-\cos 2 x}{x 	an 2 x+\frac{2 x}{3} 	an 3 x}$ Using the identity $1-\cos 2x = 2 \sin^2 x$,we get: $= \lim _

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$\frac{1}{2}$

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(B) Given,$\lim _{x ightarrow 0} \frac{1-\cos 2 x}{x 	an 2 x+\frac{2 x}{3} 	an 3 x}$ Using the identity $1-\cos 2x = 2 \sin^2 x$,we get: $= \lim _{x ightarrow 0} \frac{2 \sin^2 x}{x 	an 2 x + \frac{2 x}{3} 	an 3 x}$ Divide numerator and denominator by $x^2$: $= \lim _{x ightarrow 0} \frac{2 (\frac{\sin x}{x})^2}{\frac{	an 2 x}{x} + \frac{2}{3} \frac{	an 3 x}{x}}$ $= \lim _{x ightarrow 0} \frac{2 (\frac{\sin x}{x})^2}{2 (\frac{	an 2 x}{2 x}) + 2 (\frac{	an 3 x}{3 x})}$ Using $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x ightarrow 0} \frac{	an kx}{kx} = 1$: $= \frac{2(1)^2}{2(1) + 2(1)} = \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}$

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)}{x \tan 2 x+\frac{2 x}{3} \tan 3 x} = $

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