mathop {lim }limits_{x to a} frac{{sqrt {a + | vedclass

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Question

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

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

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(A) Given limit: $L = \mathop {\lim }\limits_{x 	o a} \frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }}$Rationalizing the numerator and denominator:$L = \mathop {\lim }\limits_{x 	o a} \frac{(\sqrt {a + 2x} - \sqrt {3x})(\sqrt {a + 2x} + \sqrt {3x})}{(\sqrt {3a + x} - 2\sqrt x)(\sqrt {3a + x} + 2\sqrt x)} 	imes \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$$L = \mathop {\lim }\limits_{x 	o a} \frac{(a + 2x - 3x)}{(3a + x - 4x)} 	imes \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$$L = \mathop {\lim }\limits_{x 	o a} \frac{a - x}{3a - 3x} 	imes \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$$L = \mathop {\lim }\limits_{x 	o a} \frac{a - x}{3(a - x)} 	imes \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$$L = \frac{1}{3} 	imes \frac{\sqrt {3a + a} + 2\sqrt a}{\sqrt {a + 2a} + \sqrt {3a}} = \frac{1}{3} 	imes \frac{2\sqrt {4a} + 2\sqrt a}{2\sqrt {3a}} = \frac{1}{3} 	imes \frac{2(2\sqrt a) + 2\sqrt a}{2\sqrt {3a}} = \frac{1}{3} 	imes \frac{6\sqrt a}{2\sqrt {3a}} = \frac{1}{3} 	imes \frac{3}{\sqrt 3} = \frac{1}{\sqrt 3} = \frac{\sqrt 3}{3}$.Wait,re-evaluating the limit: $\frac{1}{3} 	imes \frac{2\sqrt{4a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} 	imes \frac{4\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} 	imes \frac{6\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} 	imes \frac{3}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.Correction: The provided option $B$ is $\frac{2}{3\sqrt{3}}$. Let's re-check the simplification: $\frac{1}{3} 	imes \frac{2\sqrt{4a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} 	imes \frac{4\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{6\sqrt{a}}{6\sqrt{3a}} = \frac{1}{\sqrt{3}}$.Actually,$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$. The correct value is $\frac{1}{\sqrt{3}}$.

$\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }} = \dots$ (where $a \ne 0$)

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