Question

The value of $\underset{x\to b}{lim}\frac{\sqrt{x-a}-\sqrt{b-a}}{x^2-b^2}$, for $b>a$ is?

• A. $\frac{1}{4b\sqrt{a+b}}$
• B. $\frac{1}{4b\sqrt{b-a}}$
• C. $\frac{3}{4a\sqrt{a+b}}$
• D. $\frac{3}{5b\sqrt{b-a}}$

Answer 1

The correct option is B $\frac{1}{4b\sqrt{b-a}}$

We have,
$\underset{x\to b}{lim}\frac{\sqrt{x-a}-\sqrt{b-a}}{x^2-b^2}$
$=\underset{x\to b}{lim}\frac{(\sqrt{x-a}-\sqrt{b-a})(\sqrt{x-a}+\sqrt{b-a})}{(x-b)(x+b)(\sqrt{x-a}+\sqrt{b-a})}$
$=\underset{x\to b}{lim}\frac{x-b}{(x-b)(x+b)(\sqrt{x-a}+\sqrt{b-a})}$
$=\underset{x\to b}{lim}\frac{1}{(b+b)(\sqrt{x-a}+\sqrt{b-a})}$
$=\frac{1}{2b\left(2\sqrt{b-a}\right)}=\frac{1}{4b\sqrt{b-a}}$