Question

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

• A. $\frac{1}{4a}$
• B. $\frac{1}{a\sqrt{a-b}}$
• C. $\frac{1}{2a\sqrt{a-b}}$
• D. $\frac{1}{4a\sqrt{a-b}}$

Answer 1

Answer: (D)

$\frac{1}{4a\sqrt{a-b}}$

$\underset{x\to a}{lim}\frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}=\underset{x\to a}{lim}\frac{(x-b)-(a-b)}{x^2-a^2(\sqrt{x-b}+\sqrt{a-b})}=\underset{x\to a}{lim}\frac{1}{(x+a)(\sqrt{x-b}+\sqrt{a-b})}=\frac{1}{2a(\sqrt{a-b}+\sqrt{a-b})}=\frac{1}{4a\sqrt{a-b}}$
Hence, option 'D' is correct.