## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to\infty}\frac{2\sqrt x-x^{-1}}{3x-7}=0$$
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 $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$ $$A=\lim_{x\to\infty}\frac{2\sqrt x-x^{-1}}{3x-7}=\lim_{x\to\infty}\frac{2\sqrt x-\frac{1}{x}}{3x-7}$$ The highest power of $x$ in the denominator here is $x$, so we divide both numerator and denominator by $x$: $$A=\lim_{x\to\infty}\frac{\frac{2\sqrt x}{x}-\frac{1}{x^2}}{3-\frac{7}{x}}=\lim_{x\to\infty}\frac{\frac{2}{\sqrt x}-\frac{1}{x^2}}{3-\frac{7}{x}}=\lim_{x\to\infty}\frac{\frac{2}{x^{1/2}}-\frac{1}{x^2}}{3x-7}$$ $$A=\frac{0-0}{3-0}=0$$