A) \[\frac{1}{8\sqrt{3}}\]
B) \[\frac{1}{4\sqrt{3}}\]
C) 0
D) None of these
Correct Answer: A
Solution :
[a] \[\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{1+\sqrt{2+x}-3}{\left( \sqrt{1+\sqrt{2+x}}+\sqrt{3} \right)(x-2)}\] (Rationalizing) \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{2+x}-2}{\left( \sqrt{1+\sqrt{2+x}}+\sqrt{3} \right)(x-2)}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)}{\left( \sqrt{1+\sqrt{2+x}}+\sqrt{3} \right)\left( \sqrt{2+x}+2 \right)(x-2)}\] (Rationalizing) \[=\frac{1}{(2\sqrt{3})4}=\frac{1}{8\sqrt{3}}\]
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