Answer
$$\lim_{x\to\infty}\sqrt{\frac{x^2-5x}{x^3+x-2}}=0$$
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 $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$
$$A=\lim_{x\to\infty}\sqrt{\frac{x^2-5x}{x^3+x-2}}$$
The highest power of $x$ in the denominator here is $x^3$, so we divide both numerator and denominator by $x^3$: $$A=\lim_{x\to\infty}\sqrt{\frac{\frac{1}{x}-\frac{5}{x^2}}{1+\frac{1}{x^2}-\frac{2}{x^3}}}=\sqrt{\frac{\lim_{x\to\infty}\frac{1}{x}-\lim_{x\to\infty}\frac{5}{x^2}}{\lim_{x\to\infty}1+\lim_{x\to\infty}\frac{1}{x^2}-\lim_{x\to\infty}\frac{2}{x^3}}}$$
$$A=\sqrt{\frac{0-0}{1+0-0}}=\sqrt0=0$$
