lim _{x rightarrow a} left[ frac{sqrt{a+2x} - | vedclass

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(D) To evaluate the limit $L = \lim _{x \rightarrow a} \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}}$,we rationalize the numerator and the d

Vedclass · Accepted Answer

$\frac{2}{3\sqrt{3}}$

Vedclass · Answer

(D) To evaluate the limit $L = \lim _{x \rightarrow a} \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}}$,we rationalize the numerator and the denominator: $L = \lim _{x}$ ${\rightarrow a} \left( \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{a+2x} + \sqrt{3x}} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{3a+x} - 2\sqrt{x}} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$ $L = \lim _{x}$ ${\rightarrow a} \left( \frac{(a+2x) - 3x}{(3a+x) - 4x} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$ $L = \lim _{x}$ ${\rightarrow a} \left( \frac{a-x}{3a-3x} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$ $L = \lim _{x}$ ${\rightarrow a} \left( \frac{a-x}{3(a-x)} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$ $L = \frac{1}{3} \cdot \frac{\sqrt{3a+a} + 2\sqrt{a}}{\sqrt{a+2a} + \sqrt{3a}} = \frac{1}{3} \cdot \frac{\sqrt{4a} + 2\sqrt{a}}{\sqrt{3a} + \sqrt{3a}}$ $L = \frac{1}{3} \cdot \frac{2\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \cdot \frac{4\sqrt{a}}{2\sqrt{3}\sqrt{a}} = \frac{2}{3\sqrt{3}}$

$\lim _{x \rightarrow a} \left[ \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}} \right]$ is equal to

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