A) \[\frac{1}{\sqrt{3}}\]
B) \[\frac{2}{3\sqrt{3}}\]
C) \[\frac{2}{\sqrt{3}}\]
D) \[\frac{2}{3}\]
Correct Answer: B
Solution :
\[\underset{x\to a}{\mathop{\lim }}\,\,\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\] \[=\underset{x\to a}{\mathop{\lim }}\,\,\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\times \frac{\sqrt{a+2x}+\sqrt{3x}}{\sqrt{a+2x}+\sqrt{3x}}\times \frac{\sqrt{3a+x}+2\sqrt{x}}{\sqrt{3a+x}+2\sqrt{x}}\] \[=\underset{x\to a}{\mathop{\lim }}\,\frac{\sqrt{3a+x}+2\sqrt{x}}{3\,(\sqrt{a+2x}+\sqrt{3x)}}=\frac{2}{3\sqrt{3}}\]. Aliter : Apply L-Hospital?s rule.
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