lim _{x} {rightarrow 0} frac{x+2 sin x+3 tan | vedclass

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Question

(A) Let $f(x) = x+2 \sin x+3 	an x-	an ^3 x$ and $g(x) = \sqrt{x^2+2 \sin x+	an x+3}-\sqrt{\sin ^2 x-2 	an x-x+3}$.As $x ightarrow 0$,$f(0) = 0+

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$2 \sqrt{3}$

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(A) Let $f(x) = x+2 \sin x+3 	an x-	an ^3 x$ and $g(x) = \sqrt{x^2+2 \sin x+	an x+3}-\sqrt{\sin ^2 x-2 	an x-x+3}$.As $x ightarrow 0$,$f(0) = 0+0+0-0 = 0$ and $g(0) = \sqrt{3}-\sqrt{3} = 0$. This is a $\frac{0}{0}$ form.We apply $L$'Hopital's rule by differentiating the numerator and denominator.Numerator derivative: $f'(x) = 1+2 \cos x+3 \sec ^2 x-3 	an ^2 x \sec ^2 x$.At $x=0$,$f'(0) = 1+2(1)+3(1)-0 = 6$.Denominator derivative: $g'(x) = \frac{2x+2 \cos x+\sec ^2 x}{2 \sqrt{x^2+2 \sin x+	an x+3}} - \frac{2 \sin x \cos x-2 \sec ^2 x-1}{2 \sqrt{\sin ^2 x-2 	an x-x+3}}$.At $x=0$,$g'(0) = \frac{0+2+1}{2 \sqrt{3}} - \frac{0-2-1}{2 \sqrt{3}} = \frac{3}{2 \sqrt{3}} - \frac{-3}{2 \sqrt{3}} = \frac{6}{2 \sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.The limit is $\frac{f'(0)}{g'(0)} = \frac{6}{\sqrt{3}} = 2 \sqrt{3}$.

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

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