Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 139: 53

Answer

By graphing the function on the interval $-10 \leq x \leq 10$, we would guess that the horizontal asymptote is $y = 1$ $\lim\limits_{x \to \infty}f(x) = 3$ $\lim\limits_{x \to -\infty}f(x) = 3$ The horizontal asymptote is $y = 3$ The interval $-10 \leq x \leq 10$ does not give us the true indication of the horizontal asymptote because this function approaches the asymptote $y=3$ very gradually.
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Work Step by Step

$f(x) = \frac{3x^3+500x^2}{x^3+500x^2+100x+2000}$ By graphing the function on the interval $-10 \leq x \leq 10$, we would guess that the horizontal asymptote is $y = 1$ We can evaluate the limit as $x$ approaches infinity: $\lim\limits_{x \to \infty}\frac{3x^3+500x^2}{x^3+500x^2+100x+2000}$ $= \lim\limits_{x \to \infty}\frac{3x^3/x^3+500x^2/x^3}{x^3/x^3+500x^2/x^3+100x/x^3+2000/x^3}$ $= \frac{3}{1}$ $= 3$ Similarly, $\lim\limits_{x \to -\infty}f(x) = 3$ The horizontal asymptote is $y = 3$ The interval $-10 \leq x \leq 10$ does not give us the true indication of the horizontal asymptote because this function approaches the asymptote $y=3$ very gradually.
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