Answer
$$0$$
Work Step by Step
Given $$\lim_{x\to \infty} (\sqrt{x^2+3}-x)$$
Then
\begin{align*}
\lim_{x\to \infty} (\sqrt{x^2+3}-x)&=\lim_{x\to \infty} (\sqrt{x^2+3}-x)\frac{\sqrt{x^2+3}+x}{\sqrt{x^2+3}+x}\\
&=\lim_{x\to \infty} \frac{x^2+3 -x^2}{\sqrt{x^2+3}+x}\\
&=\lim_{x\to \infty} \frac{ 3 }{\sqrt{x^2+3}+x}\\
&=\lim_{x\to \infty} \frac{ 3/x}{\sqrt{x^2/x^2+3/x^2}+x/x}\\
&=\lim_{x\to \infty} \frac{ 3/x}{\sqrt{1+3/x^2}+1}\\
&=\frac{0}{1}\\
&=0
\end{align*}