Answer
$$\lim_{x\to\infty}\sqrt{\frac{8x^2-3}{2x^2+x}}=2$$
Work Step by Step
We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator.
Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$
$$A=\lim_{x\to\infty}\sqrt{\frac{8x^2-3}{2x^2+x}}$$
The highest power of $x$ in the denominator here is $x^2$, so we divide both numerator and denominator by $x^2$: $$A=\lim_{x\to\infty}\sqrt{\frac{8-\frac{3}{x^2}}{2+\frac{1}{x}}}=\sqrt{\frac{\lim_{x\to\infty}8-\lim_{x\to\infty}\frac{3}{x^2}}{\lim_{x\to\infty}2+\lim_{x\to\infty}\frac{1}{x}}}$$
$$A=\sqrt{\frac{8-0}{2+0}}=\sqrt{\frac{8}{2}}=\sqrt4=2$$
