Question

The value of ${lim}_{x\to 0}\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a+x}-\sqrt{a-x}}$ is

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

Answer 1

The correct option is

B

$\sqrt{a}$

${lim}_{x\to 0}\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a+x}-\sqrt{a-x}}$
Multiplying $(\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2})$ and $(\sqrt{a+x}+\sqrt{a-x})$ in N' and D'
$={lim}_{x\to 0}\frac{(a^2+x^2-ax-a^2-ax-a^2)(\sqrt{a+x}+\sqrt{a-x})}{(a+x+a+x)(\sqrt{x^2-ax+x^2}+\sqrt{a^2+ax+x^2})}$
$={lim}_{x\to 0}\frac{-2ax(\sqrt{a+x}+\sqrt{a-x})}{2x(\sqrt{a^2+ax+x^2}+\sqrt{a^2+ax+x^2})}$
$={lim}_{x\to 0}\frac{-2a(\sqrt{a+0}+\sqrt{a-0})}{2(\sqrt{a^2+a\left(0\right)+(0{)}^2}+\sqrt{a^2+a\left(0\right)+(0{)}^2})}$
$=\frac{-a\left(2\sqrt{a}\right)}{2a}=-\sqrt{a}$