Chọn phương án B.

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

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