Question

$\underset{x\to a}{\lim }\left(\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\right)=$

• A. $\frac{2}{3}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{3\sqrt{3}}{2}$
• D. $\frac{2}{3\sqrt{3}}$

Answer 1

The correct option is D

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

Explanation for the correct option:

Find the value of $\underset{x\to a}{\lim }\left(\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\right)$:

The given equation is,

$$\begin{gathered} I=\underset{x\to a}{\lim }\left(\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\right) \\ I=\left(\frac{\sqrt{a+2a}-\sqrt{3a}}{\sqrt{3a+a}-2\sqrt{a}}\right) \\ I=\left(\frac{\sqrt{3a}-\sqrt{3a}}{\sqrt{4a}-2\sqrt{a}}\right) \\ I=\frac{0}{0} \end{gathered}$$

Using the L' Hospitality rule.

$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

$$\begin{gathered} I=\underset{x\to a}{\lim }\left(\frac{{\displaystyle \frac{2}{2\sqrt{a+2x}}}-{\displaystyle \frac{\sqrt{3}.\sqrt{3}}{2\sqrt{3x}}}}{{\displaystyle \frac{1}{2\sqrt{3a+x}}}-2{\displaystyle \frac{1}{2\sqrt{x}}}}\right) \\ I=\underset{x\to a}{\lim }\left(\frac{{\displaystyle \frac{1}{\sqrt{a+2x}}-\frac{\sqrt{3}}{2\sqrt{x}}}}{{\displaystyle \frac{1}{2\sqrt{3a+x}}-\frac{1}{\sqrt{x}}}}\right) \\ I=\left(\frac{{\displaystyle \frac{1}{\sqrt{a+2a}}-\frac{\sqrt{3}}{2\sqrt{a}}}}{{\displaystyle \frac{1}{2\sqrt{3a+a}}-\frac{1}{\sqrt{a}}}}\right) \\ I=\left(\frac{{\displaystyle \frac{1}{\sqrt{3a}}-\frac{\sqrt{3}}{2\sqrt{a}}}}{{\displaystyle \frac{1}{4\sqrt{a}}-\frac{1}{\sqrt{a}}}}\right) \\ I=\left(\frac{{\displaystyle \frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{4}-1}}\right) \\ I=\left(\frac{{\displaystyle \frac{2-3}{2\sqrt{3}}}}{{\displaystyle -\frac{3}{4}}}\right) \\ I=\frac{-1}{2\sqrt{3}}\times -\frac{4}{3} \\ I=\frac{2}{3\sqrt{3}} \end{gathered}$$

Hence, the correct answer is option (D).