Question

$\underset{x\to 0}{lim}\left(\frac{\sqrt{a+2x}-\sqrt{3a}}{\sqrt{3a+x}}\right)$.

Answer 1

${lim}_{x\to 0}\frac{\sqrt{a+2x}-\sqrt{3a}}{\sqrt{3a+x}}$

Putting the value of x as 0

$\frac{\sqrt{a+2\times 0}-\sqrt{3a}}{\sqrt{3a+0}}$

$\frac{\sqrt{a+0}-\sqrt{3a}}{\sqrt{3a}}$

$\frac{\sqrt{a}-\sqrt{3a}}{\sqrt{3a}}$

$\frac{\sqrt{a}-\sqrt{3}\times \sqrt{a}}{\sqrt{3}\times \sqrt{a}}$

Now, taking $\sqrt{a}$ common

$\frac{\sqrt{a}(1-\sqrt{3})}{\sqrt{a}\times \sqrt{3}}$

Cancelling $\sqrt{a}$

$\frac{1-\sqrt{3}}{\sqrt{3}}$

Multiplying the numerator and denominator by $\sqrt{3}$

$\frac{1-\sqrt{3}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

Answer- $\frac{\sqrt{3}-3}{3}$