Question

$\underset{x\to a}{lim}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$ is equal to

• A. $\frac{2}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\frac{2}{3\sqrt{3}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$

$\underset{x\to a}{lim}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}=\frac{0}{0}$ form

Applying $L-$ Hospital's rule

$\underset{x\to a}{lim}\frac{\frac{1}{2\sqrt{a+2x}}\left(2\right)-\sqrt{3}.\frac{1}{2\sqrt{x}}}{\frac{1}{\sqrt{3a+x}}-\frac{2}{2\sqrt{x}}}$

$=\underset{x\to a}{lim}\frac{\frac{1}{\sqrt{a+2x}}-\frac{\sqrt{3}}{2\sqrt{x}}}{\frac{1}{\sqrt{3a+x}}-\frac{1}{\sqrt{x}}}$

$=\frac{\frac{1}{\sqrt{3}a}-\frac{\sqrt{3}}{2\sqrt{a}}}{\frac{1}{\sqrt{4a}}-\frac{1}{\sqrt{a}}}=\frac{\frac{1}{\sqrt{3}.\sqrt{a}}-\frac{\sqrt{3}}{2\sqrt{a}}}{\frac{1}{2\sqrt{a}}-\frac{1}{\sqrt{a}}}$

$=\frac{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=\frac{\frac{2-3}{2\sqrt{3}}}{\frac{-1}{2}}=\frac{-1}{\sqrt{3}(-1)}=\frac{1}{\sqrt{3}}$