$\underset{x\to \infty }{lim}(\sqrt{x^2+ax+a^2}-\sqrt{x^2+a^2})=$

The correct option is D: $\frac{a}{2}$
$\sqrt{x^2+ax+a^2}-\sqrt{x^2+a^2}$

$=\frac{(x^2+ax+a^2)-(x^2+a^2)}{\sqrt{x^2+ax+a^2}+\sqrt{x^2+a^2}}$

$=\frac{ax}{\sqrt{x^2+ax+a^2}+\sqrt{x^2+a^2}}$

$=\frac{ax}{\sqrt{x^2(1+\frac{a}{x}+\frac{a^2}{x^2})}+\sqrt{x^2(1+\frac{a^2}{x^2}})}$

$=\frac{a}{\sqrt{1+\frac{a}{x}+\frac{a^2}{x^2}}+\sqrt{1+\frac{a^2}{x^2}}}$

Then

${lim}_{x\to \infty }(\sqrt{x^2+ax+a^2}-\sqrt{x^2+a^2})$

$={lim}_{x\to \infty }\left(\frac{a}{\sqrt{1+\frac{a}{x}+\frac{a^2}{x^2}}+\sqrt{1+\frac{a^2}{x^2}}}\right)$

$=\frac{a}{\sqrt{1+0+0}+\sqrt{1+0}}$ [$\because {lim}_{x\to \infty },\frac{1}{x}\to 0$]

$=\frac{a}{2}$