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Evaluate $\lim\limits_{\large x\to a}\large\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt x}$, where $a\neq 0$

$\begin{array}{1 1}(a)\;\large\frac{2}{\sqrt 3}\\(b)\;\large\frac{2}{3\sqrt 3}\\(c)\;2\\(d)\;\large\frac{3}{2}\end{array}$

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$\lim\limits_{\large x\to a}\large\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt x}$
$\Rightarrow \lim\limits_{x\to a}\large\frac{(\sqrt{a+2x}-\sqrt{3x})(\sqrt{a+2x}+\sqrt{3x})(\sqrt{3a+x}-2\sqrt{x})}{(\sqrt{3a+x}-2\sqrt{x})(\sqrt{3a+x}+2\sqrt{x})(\sqrt{a+2x}+\sqrt{3x})}$
$\Rightarrow \lim\limits_{\large x\to a}\large\frac{(a+2x-3x)(\sqrt{3a+x}+2\sqrt x)}{(3a+x-4x)(\sqrt{a+2x}+\sqrt{3x})}$
$\Rightarrow \lim\limits_{\large x\to a}\large\frac{(a-x)(\sqrt{3a+x}+2\sqrt x)}{3(a-x)(\sqrt{a+2x}+\sqrt{3x})}$
$\Rightarrow \lim\limits_{x\to a}\large\frac{\sqrt{3a+x}+2\sqrt x}{3(\sqrt{a+2x}+\sqrt{3x})}$
$\Rightarrow \large\frac{\sqrt{3a+a}+2\sqrt a}{3(\sqrt{a+2a}+\sqrt {3a})}$
$\Rightarrow \large\frac{2\sqrt{a}+2\sqrt a}{3\times 2\sqrt{3a}}$
$\Rightarrow \large\frac{4\sqrt{a}}{3\times 2\sqrt{3a}}$
$\Rightarrow \large\frac{2}{3 \sqrt{3}}$
Hence (b) is the correct answer.
answered Jan 4, 2014 by sreemathi.v
 

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