Chọn phương án B.

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

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