Answer
$\dfrac{1}{2}$
Work Step by Step
When $\lim\limits_{x \to \infty} f(x)$ exists, then the sequence $a_n=f(n)$ converges to the same limit, so we have:
$ \lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} f(x)$
Now, $ \lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} x (\sqrt {x^2+1} -x) \\=\lim\limits_{x \to \infty} x (\sqrt {x^2+1} -x) \times \dfrac{(\sqrt {x^2+1} +x)}{(\sqrt {x^2+1} -x)} \\=\lim\limits_{x \to \infty} \dfrac{x}{(\sqrt {x^2+1}+x)}$
We see that this limit shows the indefinite form of $\dfrac{\infty}{\infty}$, so we apply L'Hopital's rule.
Thus, $ \lim\limits_{n \to \infty} a_n =\lim\limits_{x \to \infty} \dfrac{1}{\dfrac{x}{\sqrt {x^2+1}}+1} \\=\lim\limits_{x \to \infty} \dfrac{1}{\sqrt {\dfrac{1}{1+\dfrac{1}{x^2}}}+1} \\= \dfrac{1}{\sqrt {\dfrac{ \lim\limits_{x \to \infty} 1}{\lim\limits_{x \to \infty} [1+\dfrac{1}{x^2}}}+1]} \\=\dfrac{1}{\sqrt {\frac{1}{1+0}}+1}\\=\dfrac{1}{2}$
