Question

If $\underset{x\to a}{lim}\frac{2x-\sqrt{x^2+{3a}^2}}{\sqrt{x+a}-\sqrt{2a}}=\sqrt{2}$ (where $a\in R^{+})$ then a is equal to

Answer 1

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

$Theobjectiveistofinda=?{lim}_{x\to a}\frac{\left[4x^2-\left(x^2+3a^2\right)\right]\left[\sqrt{x+a}+\sqrt{2a}\right]}{\sqrt{x+a}-\sqrt{2a}}=\sqrt{2}{lim}_{x\to a}\left[\frac{3\left(x-a\right)\left(x+a\right)}{\left(x+a\right)}\right]\left(\frac{2\sqrt{2a}}{4a}\right)=\sqrt{2}\frac{6a\times 2\sqrt{2a}}{4a}=\sqrt{2}3\sqrt{2a}=\sqrt{2}3\sqrt{a}=1a=\frac{1}{9}$

Hence, this is the answer.