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

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(C) Let $L = \lim _{x ightarrow 0} \frac{2 	an x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 	an x+1}-\sqrt{3 	an ^2 x+\sin x+1}}$.Rationalizing the denomina

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$6$

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(C) Let $L = \lim _{x ightarrow 0} \frac{2 	an x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 	an x+1}-\sqrt{3 	an ^2 x+\sin x+1}}$.Rationalizing the denominator:$L = \lim _{x ightarrow 0} \frac{(2 	an x + \cos x - 1 + x)(\sqrt{4 \sin ^2 x + 2 	an x + 1} + \sqrt{3 	an ^2 x + \sin x + 1})}{(4 \sin ^2 x + 2 	an x + 1) - (3 	an ^2 x + \sin x + 1)}$.As $x ightarrow 0$,the term $(\sqrt{4 \sin ^2 x + 2 	an x + 1} + \sqrt{3 	an ^2 x + \sin x + 1}) ightarrow \sqrt{1} + \sqrt{1} = 2$.So,$L = 2 \lim _{x ightarrow 0} \frac{2 	an x + x + (\cos x - 1)}{4 \sin ^2 x - 3 	an ^2 x + 2 	an x - \sin x}$.Using small angle approximations $	an x \approx x$,$\sin x \approx x$,$\cos x - 1 \approx -\frac{x^2}{2}$:$L = 2 \lim _{x ightarrow 0} \frac{2x + x - \frac{x^2}{2}}{4x^2 - 3x^2 + 2x - x} = 2 \lim _{x ightarrow 0} \frac{3x - \frac{x^2}{2}}{x^2 + x} = 2 \lim _{x ightarrow 0} \frac{3 - \frac{x}{2}}{x + 1} = 2 	imes \frac{3}{1} = 6$.

$\lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}-\sqrt{3 \tan ^2 x+\sin x+1}} = $

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