The value of mathop {lim }limits_{x to 2} fra | vedclass

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Question

(A) Let $L = \mathop {\lim }\limits_{x 	o 2} \frac{{\sqrt {1 + \sqrt {2 + x} } - \sqrt 3 }}{{x - 2}}$.Rationalizing the numerator:$L = \mathop {\lim

Vedclass · Accepted Answer

$\frac{1}{8\sqrt{3}}$

Vedclass · Answer

(A) Let $L = \mathop {\lim }\limits_{x 	o 2} \frac{{\sqrt {1 + \sqrt {2 + x} } - \sqrt 3 }}{{x - 2}}$.Rationalizing the numerator:$L = \mathop {\lim }\limits_{x 	o 2} \frac{(\sqrt {1 + \sqrt {2 + x} } - \sqrt 3)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)}$$L = \mathop {\lim }\limits_{x 	o 2} \frac{1 + \sqrt {2 + x} - 3}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)}$$L = \mathop {\lim }\limits_{x 	o 2} \frac{\sqrt {2 + x} - 2}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)}$Rationalizing the numerator again:$L = \mathop {\lim }\limits_{x 	o 2} \frac{(\sqrt {2 + x} - 2)(\sqrt {2 + x} + 2)}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)(\sqrt {2 + x} + 2)}$$L = \mathop {\lim }\limits_{x 	o 2} \frac{2 + x - 4}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)(\sqrt {2 + x} + 2)}$$L = \mathop {\lim }\limits_{x 	o 2} \frac{x - 2}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)(\sqrt {2 + x} + 2)}$$L = \frac{1}{(\sqrt {1 + \sqrt {4} } + \sqrt 3)(\sqrt {4} + 2)} = \frac{1}{(\sqrt 3 + \sqrt 3)(4)} = \frac{1}{2\sqrt 3 	imes 4} = \frac{1}{8\sqrt 3}$.

The value of $\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {1 + \sqrt {2 + x} } - \sqrt 3 }}{{x - 2}}$ is

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