Chọn phương án D.

\(\begin{aligned}
&\lim\limits_{x\to0^+}\dfrac{\sqrt{x^2+x}-\sqrt{x}}{x^2}\\
=&\lim\limits_{x\to0^+}\dfrac{\left(\sqrt{x^2+x}-\sqrt{x}\right)\left(\sqrt{x^2+x}+\sqrt{x}\right)}{x^2\left(\sqrt{x^2+x}+\sqrt{x}\right)}\\
=&\lim\limits_{x\to0^+}\dfrac{\left(x^2+x\right)-x}{x^2\left(\sqrt{x^2+x}+\sqrt{x}\right)}\\
=&\lim\limits_{x\to0^+}\dfrac{x^2}{x^2\left(\sqrt{x^2+x}+\sqrt{x}\right)}\\
=&\lim\limits_{x\to0^+}\dfrac{1}{\sqrt{x^2+x}+\sqrt{x}}\\
=&+\infty.
\end{aligned}\)

Vì \(\begin{cases}
\lim\limits_{x\to0^+}1&=1>0\\
\lim\limits_{x\to0^+}\left(\sqrt{x^2+x}+\sqrt{x}\right)&=0\\
\sqrt{x^2+x}+\sqrt{x}>0&\text{khi }x\to0^+.
\end{cases}\)