Question

$\underset{x\to 0}{lim}\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a+x}-\sqrt{a-x}}$ is equal to (a > 0)

• A. $\sqrt{a}$
• B. $-\sqrt{a}$
• C. $a\sqrt{a}$
• D. $-a\sqrt{a}$

Answer 1

Answer: (B)

$-\sqrt{a}$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{(\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2})(\sqrt{a+x}+\sqrt{a-x})}{(a+x)-(a-x)}$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{2\sqrt{a}}{2x}(\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2})$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{\sqrt{a}}{x}\left(\frac{(a^2-ax+x^2)-(a^2+ax+x^2)}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}\right)$

$\Rightarrow L=\underset{x\to 0}{lim}\frac{\sqrt{a}}{x}\times \frac{(-2ax)}{2\sqrt{a}}=-\sqrt{a}$

$\Rightarrow \left(B\right)$.