Question

The value of ${lim}_{x\to 2a}\frac{\sqrt{x-2a}+\sqrt{x}-\sqrt{2a}}{\sqrt{x^2-4a^2}}$ is

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

Answer 1

The correct option is B $\frac{1}{2\sqrt{a}}$

put $x=2a$

we get $\frac{0}{0}$ form

apply L'hospital rule

differentiate the function seperately(numerator and denominator)

we get

$\frac{d}{dx}\frac{\sqrt{x-2a}+\sqrt{x}-\sqrt{2a}}{\sqrt{x^2-4a^2}}$

numerator is $\frac{1}{2\sqrt{x-2a}}+\frac{1}{\sqrt{x}}$

denominator is $\frac{x}{\sqrt{x^2-4a^2}}$

now combine the function

$\frac{numerator}{denominator}=\frac{1}{2}(\sqrt{x}+\sqrt{x-2a})\frac{\sqrt{x+2a}}{x}$

now put x=2a

we get $\frac{1}{2\sqrt{a}}$