lim _{x rightarrow 0} frac{sqrt{11+|x|-6 sqrt | vedclass

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(D) Let $f(x) = \frac{\sqrt{11+|x|-6 \sqrt{2+|x|}}}{6-2 \sqrt{2+|x|}}$.We observe that $11+|x|-6 \sqrt{2+|x|} = 9 + 2 + |x| - 6\sqrt{2+|x|} = 3^2 + (\

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$\frac{1}{2}$

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(D) Let $f(x) = \frac{\sqrt{11+|x|-6 \sqrt{2+|x|}}}{6-2 \sqrt{2+|x|}}$.We observe that $11+|x|-6 \sqrt{2+|x|} = 9 + 2 + |x| - 6\sqrt{2+|x|} = 3^2 + (\sqrt{2+|x|})^2 - 2(3)(\sqrt{2+|x|}) = (3-\sqrt{2+|x|})^2$.Substituting this into the limit expression:$\lim _{x \rightarrow 0} \frac{\sqrt{(3-\sqrt{2+|x|})^2}}{2(3-\sqrt{2+|x|})}$$= \lim _{x \rightarrow 0} \frac{|3-\sqrt{2+|x|}|}{2(3-\sqrt{2+|x|})}$.As $x \rightarrow 0$,$\sqrt{2+|x|} \rightarrow \sqrt{2} \approx 1.414$,so $3-\sqrt{2+|x|} > 0$.Thus,$|3-\sqrt{2+|x|}| = 3-\sqrt{2+|x|}$.$= \lim _{x \rightarrow 0} \frac{3-\sqrt{2+|x|}}{2(3-\sqrt{2+|x|})} = \frac{1}{2}$.

$\lim _{x \rightarrow 0} \frac{\sqrt{11+|x|-6 \sqrt{2+|x|}}}{6-2 \sqrt{2+|x|}} = $

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