# Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 107: 23

$$\lim_{x\to\infty}\sqrt{\frac{8x^2-3}{2x^2+x}}=2$$

#### 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 $x^2$: $$A=\lim_{x\to\infty}\sqrt{\frac{8-\frac{3}{x^2}}{2+\frac{1}{x}}}=\sqrt{\frac{\lim_{x\to\infty}8-\lim_{x\to\infty}\frac{3}{x^2}}{\lim_{x\to\infty}2+\lim_{x\to\infty}\frac{1}{x}}}$$ $$A=\sqrt{\frac{8-0}{2+0}}=\sqrt{\frac{8}{2}}=\sqrt4=2$$

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