Question

Evaluate: $\underset{x\to 2}{lim}\frac{\sqrt{x-2}+\sqrt{x}-\sqrt{2}}{\sqrt{x^2-4}}$

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

Answer: (C)

$\frac{1}{2}$

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

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

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

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

$=\underset{x\to 2}{lim}(\frac{1}{\sqrt{2+2}}+\frac{\sqrt{4-4}}{2\left(2\right)\sqrt{2}})$

$=\frac{1}{2}$