Question

$\underset{x\to a}{Lt}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$.

Answer 1

$\begin{array}{c}\frac{0}{0}from\\ u\sin gL.hospitalrule\\ {lim}_{x\to a}=\frac{\frac{2}{2\sqrt{a+2x}}-\frac{3}{\sqrt{3x}}}{\frac{1}{2\sqrt{3a+x}}-\frac{2}{2\sqrt{x}}}\\ {lim}_{x\to a}=\frac{\frac{1}{\sqrt{a+2x}}-\frac{\sqrt{3}}{\sqrt{x}}}{\frac{1}{2\sqrt{3a+x}-\frac{1}{\sqrt{x}}}}={lim}_{x\to a}\frac{\frac{1}{\sqrt{a+2a}}-\frac{\sqrt{3}}{\sqrt{a}}}{\frac{1}{2\sqrt{3a+a}}-\frac{1}{\sqrt{a}}}=\frac{\left(d\frac{\sqrt{a}-\sqrt{3a+6a}}{\sqrt{a}x\sqrt{3a}}\right)}{\left(\frac{\sqrt{a}-2\sqrt{4a}}{2\sqrt{a}.\sqrt{4a}}\right)}\\ \frac{\sqrt{a}(1-3)}{\sqrt{3a}}\times \frac{4a}{\sqrt{a(1-4})}=\frac{-2}{-3}\times d\frac{4}{\sqrt{3}}=\frac{8}{3\sqrt{3}}\end{array}$