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Q. $\displaystyle \lim_{x \to a} \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}} $ = ______

KCETKCET 2011Limits and Derivatives

Solution:

$\displaystyle \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$
$=\displaystyle \lim _{x \rightarrow a} \frac{\sqrt{3 a+ x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}} \times \frac{a+2 x-3 x}{3 a+x-4 x}$
$=\displaystyle \lim _{x \rightarrow a} \frac{\sqrt{3 a +x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}} \times \frac{a-x}{3 a-3 x}$
$=\displaystyle \lim _{x \rightarrow a} \cdot \frac{1}{3}\left\{\frac{\sqrt{3 a +x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right\} \cdot\left(\frac{a-x}{a-x}\right)$
$=\frac{1}{3}\left\{\frac{\sqrt{4 a}+2 \sqrt{a}}{\sqrt{a+2 a}+\sqrt{3 a}}\right\}$
$=\frac{1}{3}\left(\frac{2 \sqrt{a}+2 \sqrt{a}}{\sqrt{3 a}+\sqrt{3 a}}\right)$
$=\frac{1}{3} \cdot \frac{4 \sqrt{a}}{2 \sqrt{3} \cdot \sqrt{a}}=\frac{2}{3 \sqrt{3}}$