Single Variable Calculus

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Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 40

Expert Answers

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a.) Illustrate the graph of $\displaystyle f(x) = \frac{\sqrt{2x^2 + 1} }{3x - 5}$. How many horizontal and vertical asymptotes do you observe? Use the graph to estimate the values of the limits

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5}$ and $\displaystyle \lim_{x \to - \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5}$

Based from the graph, there are two horizontal asymptotes $y = -0.5$ and $y = 0.5$ and the vertical asymptote is $\displaystyle x = 1.6, \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1} }{3x - 5} = 0.5$ and $\lim_{x \to \infty} \frac{\sqrt{2x^2 + 1} }{3x - 5} = -0.5$.

b.) Find values of $f(x)$ to give estimates of the limits in part (a).

$ \begin{array}{|c|c|} \hline\\ \text{Values of $f(x)$ as $x$ approaches $\infty$}\\ x & f(x) \\ 10 & 0.5671 \\ 100 & 0.4794 \\ 1000 & 0.4722 \\ 10000 & 0.4715 \\ 100000 & 0.4714 \\ 1000000 & 0.4714 \\ \hline \text{Values of $f(x)$ as $x$ aprroaches $- \infty$} & \\ x & f(x)\\ -10 & -0.4051\\ -100 & -0.4637\\ -1000 & -0.4706\\ -10000 & -0.4713 \\ -10000 & -0.4714\\ -1000000 & -0.4714\\ \hline \end{array} $

Based from the table, both values approaches $\pm 0.4714$ or close to $\pm 0.5$.

c.) Find the exact values of the limits in part (a)

$ \begin{equation} \begin{aligned} \lim_{x \to \infty} \frac{\sqrt{2x^2 + 1}}{3x - 5} \cdot \frac{\displaystyle \frac{1}{\sqrt{x^2}}}{\frac{1}{x}} =& \lim_{x \to \infty} \frac{\displaystyle \sqrt{\frac{2 \cancel{x^2}}{\cancel{x^2}}} + \frac{1}{x^2} }{\displaystyle \frac{3 \cancel{x}}{\cancel{x}} - \frac{5}{x}} \\ \\ =& \lim_{x \to \infty} \frac{\displaystyle \sqrt{2 + \frac{1}{x^2}}}{\displaystyle 3 - \frac{5}{x}} \\ \\ =& \frac{ \displaystyle \lim_{x \to \infty} \sqrt{2 + \frac{1}{x^2}} }{\displaystyle \lim_{x \to \infty} 3 - \frac{5}{x} } \\ \\ =& \frac{\displaystyle \sqrt{2 + \lim_{x \to \infty} \frac{1}{x^2} } }{\displaystyle 3 - \lim_{x \to \infty} \frac{5}{x}} \\ \\ =& \frac{\sqrt{2 + 0}}{3 - 0} \\ \\ =& \frac{\sqrt{2}}{3} \text{ or } 0.4714 \\ \\ \lim_{x \to - \infty} \frac{\sqrt{2x^2 + 1}}{-(3x - 5)} \cdot \frac{\displaystyle \frac{1}{\sqrt{x^2}}}{\displaystyle \frac{1}{x}} =& \lim_{x \to - \infty} \frac{\displaystyle \sqrt{\frac{2 \cancel{x^2}}{\cancel{x^2}}} + \frac{1}{x^2} }{\displaystyle \frac{5}{x} - \frac{3 \cancel{x}}{\cancel{x}} } \\ \\ =& \lim_{x \to - \infty} \frac{\displaystyle \sqrt{2 + \frac{1}{x^2}}}{\displaystyle \frac{5}{x } - 3} \\ \\ =& \frac{\displaystyle \lim_{x \to - \infty} \sqrt{2 + \frac{1}{x^2}} }{\displaystyle \lim_{x \to - \infty} \frac{5}{x} - 3} \\ \\ =& \frac{\displaystyle \sqrt{2 + \lim_{x \to - \infty} \frac{1}{x^2}}}{\displaystyle \lim_{x \to - \infty} \frac{5}{x} - 3} \\ \\ =& \frac{\sqrt{2 + 0}}{0 - 3} \\ \\ =& \frac{\sqrt{2}}{- 3} \\ \\ =& \frac{- \sqrt{2}}{3} \text{ or } -0.4714 \end{aligned} \end{equation} $

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