Answer
$2$
Work Step by Step
\begin{align*}
\lim_{x\to \infty} \sqrt{\frac{8x^2-3}{2x^2+2 }}&=\lim_{x\to \infty} \sqrt{\frac{8x^2/x^2-3/x^2}{2x^2/x^2+2/x^2 }}\\
&=\lim_{x\to \infty} \sqrt{\frac{8-3/x^2}{2+2/x^2 }}\\
&=\sqrt{\frac{\lim_{x\to \infty}(8)-\lim_{x\to \infty}(3/x^2)}{\lim_{x\to \infty}(2)+\lim_{x\to \infty}(2/x^2) }}\\
&=\sqrt{\frac{8-0}{2+0 }}\\
&=2
\end{align*}