Question

If ${lim}_{x\to a}\frac{2x-\sqrt{x^2+3a^2}}{\sqrt{x+a}-\sqrt{2a}}=\sqrt{2}$(where $aϵR^{+}),$ then a is equal to -

• A. $\frac{1}{3}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{3\sqrt{2}}$
• D. $\frac{1}{9}$

Answer 1

The correct option is D $\frac{1}{9}$

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

$\%$ form L Hospital rule

$\underset{x\to a}{lim}\frac{2-\frac{2x}{2\sqrt{x^2+3a^2}}}{\frac{1}{2\sqrt{x+a}}-0}=\sqrt{2}$

$\Rightarrow \frac{2-\frac{2a}{2\sqrt{a^2+3a^2}}}{\frac{1}{2\sqrt{a+a}}-0}=\sqrt{2}$

$\Rightarrow \frac{2-\frac{a}{2a}}{\frac{1}{2\sqrt{2a}}}=\sqrt{2}$

$\Rightarrow \frac{3}{2}\times 2\sqrt{2a}=\sqrt{2}$

$\Rightarrow 3\sqrt{a}=1$

$\Rightarrow a=\frac{1}{9}$