mathop {lim }limits_{x to 3} frac{{sqrt {3x} | vedclass

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Question

(B) Let $A = \mathop {\lim }\limits_{x 	o 3} \frac{{\sqrt {3x} - 3}}{{\sqrt {2x - 4} - \sqrt 2 }}$.Rationalizing the numerator and the denominator:$A

Vedclass · Accepted Answer

$\frac{1}{{\sqrt 2 }}$

Vedclass · Answer

(B) Let $A = \mathop {\lim }\limits_{x 	o 3} \frac{{\sqrt {3x} - 3}}{{\sqrt {2x - 4} - \sqrt 2 }}$.Rationalizing the numerator and the denominator:$A = \mathop {\lim }\limits_{x 	o 3} \frac{(\sqrt{3x} - 3)(\sqrt{3x} + 3)(\sqrt{2x - 4} + \sqrt{2})}{(\sqrt{2x - 4} - \sqrt{2})(\sqrt{2x - 4} + \sqrt{2})(\sqrt{3x} + 3)}$$A = \mathop {\lim }\limits_{x 	o 3} \frac{(3x - 9)(\sqrt{2x - 4} + \sqrt{2})}{(2x - 4 - 2)(\sqrt{3x} + 3)}$$A = \mathop {\lim }\limits_{x 	o 3} \frac{3(x - 3)(\sqrt{2x - 4} + \sqrt{2})}{2(x - 3)(\sqrt{3x} + 3)}$Canceling $(x - 3)$:$A = \frac{3}{2} 	imes \frac{\sqrt{2(3) - 4} + \sqrt{2}}{\sqrt{3(3)} + 3}$$A = \frac{3}{2} 	imes \frac{\sqrt{2} + \sqrt{2}}{3 + 3} = \frac{3}{2} 	imes \frac{2\sqrt{2}}{6} = \frac{3}{2} 	imes \frac{\sqrt{2}}{3} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

$\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {3x} - 3}}{{\sqrt {2x - 4} - \sqrt 2 }}$ is equal to

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