Question

$\underset{x\to 2}{lim}\frac{\sqrt{x-2}+\sqrt{x}-\sqrt{2}}{\sqrt{x^2-4}}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$\underset{x\to 2}{lim}\frac{\sqrt{x-2}+\sqrt{x}-\sqrt{2}}{\sqrt{x^2-4}}$
$=\underset{x\to 2}{lim}(\frac{\sqrt{x-2}}{\sqrt{x+2}\sqrt{x-2}}+\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x^2-4}})$
On rationalisation -
$=\underset{x\to 2}{lim}(\frac{1}{\sqrt{x+2}}+\frac{x-2}{\sqrt{x^2-4}(\sqrt{x}+\sqrt{2})})$
$=\underset{x\to 2}{lim}\frac{1}{\sqrt{x+2}}+\underset{x\to 2}{lim}\sqrt{\frac{x-2}{x+2}}\times \frac{1}{\sqrt{x}+\sqrt{2}}$
$=\frac{1}{2}$