Question

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

(a) a

(b) $\sqrt{a}$

(c) −a

(d) $-\sqrt{a}$

Answer 1

(d) $-\sqrt{a}$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a+x}-\sqrt{a-x}} \\ \mathrm{Multiplying}\left(\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}\right)\text{and}\left(\sqrt{a+x}+\sqrt{a-x}\right)\mathrm{in}{\mathrm{N}}^r\mathrm{and}{\mathrm{D}}^r: \\ \underset{x\to 0}{\lim }\frac{\left(a^2+x^2-ax-a^2-ax-x^2\right)\left(\sqrt{a+x}+\sqrt{a-x}\right)}{\left(a+x-a+x\right)\left(\sqrt{x^2-ax+x^2}+\sqrt{a^2+ax+x^2}\right)} \\ \underset{x\to 0}{\lim }\frac{-2ax\left(\sqrt{a+x}+\sqrt{a-x}\right)}{2x\left(\sqrt{a^2+ax+x^2}+\sqrt{a^2+ax+x^2}\right)} \\ =\frac{-a\left(2\sqrt{a}\right)}{2a}=-\sqrt{a} \end{gathered}$$