Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 83

Answer

Both techniques yield $$\lim_{x\to\infty}\frac{2x^3-x^2+1}{5x^3+2x}=\frac{2}{5}.$$

Work Step by Step

1) Standard method (identify the highest power of $x$ and pull it in front of the parentheses) $$\lim_{x\to\infty}\frac{2x^3-x^2+1}{5x^3+2x}=\lim_{x\to\infty}\frac{x^3\left(\frac{2x^3}{x^3}-\frac{x^2}{x^3}+\frac{1}{x^3}\right)}{x^3\left(\frac{5x^3}{x^3}-\frac{2x}{x^3}\right)}=\lim_{x\to\infty}\frac{2-\frac{1}{x}+\frac{1}{x^3}}{5+\frac{2}{x^2}}=\left[\frac{2-0+0}{5+0}\right]=\frac{2}{5}.$$ 2) L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to\infty}\frac{2x^3-x^2+1}{5x^3+2x}=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(2x^3-x^2+1)'}{(5x^3+2x)'}=\lim_{x\to\infty}\frac{6x^2-2x}{15x^2+2}=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(6x^2-2x)'}{(15x^2+2)'}=\lim_{x\to\infty}\frac{12x-2}{30x}=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(12x-2)'}{(30x)'}=\lim_{x\to\infty}\frac{12}{30}=\frac{2}{5}.$$

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