Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 139: 54

(a) When we use the zoom function on a graphing calculator, we could estimate that: $\lim\limits_{x \to \infty}f(x) = 0.47$ $\lim\limits_{x \to -\infty}f(x) = -0.47$ (b) We could estimate that: $\lim\limits_{x \to \infty}f(x) = 0.4714$ $\lim\limits_{x \to -\infty}f(x) = -0.4714$ (c) $\lim\limits_{x \to \infty}\frac{\sqrt{2x^2+1}}{3x-5}=\frac{\sqrt{2}}{3}$ $\lim\limits_{x \to -\infty}\frac{\sqrt{2x^2+1}}{3x-5} = -\frac{\sqrt{2}}{3}$

(a) $f(x) = \frac{\sqrt{2x^2+1}}{3x-5}$ On the graph, we can see two horizontal asymptotes. When we use the zoom function on a graphing calculator, we could estimate that: $\lim\limits_{x \to \infty}f(x) = 0.47$ $\lim\limits_{x \to -\infty}f(x) = -0.47$ (b) We can evaluate $f(x)$ for increasing values of $x$: $f(10) = \frac{\sqrt{2(10)^2+1}}{3(10)-5} = 0.5671$ $f(100) = \frac{\sqrt{2(100)^2+1}}{3(100)-5} = 0.4794$ $f(1000) = \frac{\sqrt{2(1000)^2+1}}{3(1000)-5} = 0.4722$ $f(10,000) = \frac{\sqrt{2(10,000)^2+1}}{3(10,000)-5} = 0.4715$ $f(100,000) = \frac{\sqrt{2(100,000)^2+1}}{3(100,000)-5} = 0.4714$ $f(1,000,000) = \frac{\sqrt{2(1,000,000)^2+1}}{3(1,000,000)-5} = 0.4714$ We could estimate that: $\lim\limits_{x \to \infty}f(x) = 0.4714$ We can evaluate $f(x)$ for decreasing values of $x$: $f(-10) = \frac{\sqrt{2(-10)^2+1}}{3(-10)-5} = -0.4051$ $f(-100) = \frac{\sqrt{2(-100)^2+1}}{3(-100)-5} = -0.4637$ $f(-1000) = \frac{\sqrt{2(-1000)^2+1}}{3(-1000)-5} = -0.4706$ $f(-10,000) = \frac{\sqrt{2(-10,000)^2+1}}{3(-10,000)-5} = -0.4713$ $f(-100,000) = \frac{\sqrt{2(-100,000)^2+1}}{3(-100,000)-5} = -0.4714$ $f(-1,000,000) = \frac{\sqrt{2(-1,000,000)^2+1}}{3(-1,000,000)-5} = -0.4714$ We could estimate that: $\lim\limits_{x \to -\infty}f(x) = -0.4714$ (c) We can calculate $\lim\limits_{x \to \infty}f(x)$: $\lim\limits_{x \to \infty}\frac{\sqrt{2x^2+1}}{3x-5}$ $=\lim\limits_{x \to \infty}\frac{(\sqrt{2x^2+1})(\frac{1}{x})}{(3x-5)(\frac{1}{x})}$ $=\lim\limits_{x \to \infty}\frac{\sqrt{2x^2/x^2+1/x^2}}{3x/x-5/x}$ $=\lim\limits_{x \to \infty}\frac{\sqrt{2+1/x^2}}{3-5/x}$ $=\frac{\sqrt{2+0}}{3-0}$ $=\frac{\sqrt{2}}{3}$ We can calculate $\lim\limits_{x \to -\infty}f(x)$: $\lim\limits_{x \to -\infty}\frac{\sqrt{2x^2+1}}{3x-5}$ $=\lim\limits_{x \to -\infty}\frac{(\sqrt{2x^2+1})(\frac{1}{x})}{(3x-5)(\frac{1}{x})}$ $=\lim\limits_{x \to -\infty}\frac{-\sqrt{2x^2/x^2+1/x^2}}{3x/x-5/x}$ $=\lim\limits_{x \to -\infty}\frac{-\sqrt{2+1/x^2}}{3-5/x}$ $=\frac{-\sqrt{2+0}}{3-0}$ $=-\frac{\sqrt{2}}{3}$