Chọn phương án B.

\(\begin{aligned}
&\lim\limits_{x\to+\infty}\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+2}\right)\\
=&\lim\limits_{x\to+\infty}\left(\sqrt[3]{x^3\left(3-\dfrac{1}{x^3}\right)}+\sqrt{x^2\left(1+\dfrac{2}{x^2}\right)}\right)\\
=&\lim\limits_{x\to+\infty}\left(x\sqrt[3]{3-\dfrac{1}{x^3}}+|x|\sqrt{1+\dfrac{2}{x^2}}\right)\\
=&\lim\limits_{x\to+\infty}\left(x\sqrt[3]{3-\dfrac{1}{x^3}}+x\sqrt{1+\dfrac{2}{x^2}}\right)\\
=&\lim\limits_{x\to+\infty}x\left(\sqrt[3]{3-\dfrac{1}{x^3}}+\sqrt{1+\dfrac{2}{x^2}}\right)\\
=&+\infty.
\end{aligned}\)

Vì \(\begin{cases}
\lim\limits_{x\to+\infty}x&=+\infty\\
\lim\limits_{x\to+\infty}\left(\sqrt[3]{3-\dfrac{1}{x^3}}+\sqrt{1+\dfrac{2}{x^2}}\right)&=\sqrt[3]{3}+1>0.
\end{cases}\)