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)

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

We have,

$\underset{x\to a}{lim}\frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}$

This is the $\frac{0}{0}$ form.

Apply L-hospital rule,

${lim}_{x\to a}\frac{\frac{1}{2\sqrt{x-b}}-0}{2x-0}$

${lim}_{x\to a}\frac{1}{4x\sqrt{x-b}}$

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

Hence, this is the answer.