A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) \[-\frac{3}{2}\]
D) \[\sqrt{3}\]
Correct Answer: A
Solution :
\[=\underset{x\to \infty }{\mathop{\lim }}\,\,\sqrt{3{{x}^{2}}+\sqrt{3{{x}^{2}}+\sqrt{3{{x}^{2}}}}}-\sqrt{3{{x}^{2}}}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\,\frac{3{{x}^{2}}+\sqrt{3{{x}^{2}}+\sqrt{3{{x}^{2}}}}-3{{x}^{2}}}{\sqrt{3{{x}^{2}}+\sqrt{3{{x}^{2}}+\sqrt{3{{x}^{2}}}}}+\sqrt{3{{x}^{2}}}}\] (by rationalization) \[=\underset{x\to \infty }{\mathop{\lim }}\,\,\frac{\sqrt{3+\sqrt{\frac{3}{{{x}^{2}}}}}}{\sqrt{3+\sqrt{\frac{3}{{{x}^{2}}}}+\sqrt{\frac{3}{{{x}^{4}}}}+\sqrt{3}}}\] \[=\frac{\sqrt{3}}{\sqrt{3}+\sqrt{3}}=\frac{1}{2}\]
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