University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Practice Exercises - Page 111: 41

Answer

$$\lim_{x\to\infty}\frac{2x+3}{5x+7}=\frac{2}{5}$$

Work Step by Step

*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=0$ $$A=\lim_{x\to\infty}\frac{2x+3}{5x+7}$$ Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$ in this case: $$A=\lim_{x\to\infty}\frac{2+\frac{3}{x}}{5+\frac{7}{x}}$$ $$A=\frac{2+0}{5+0}=\frac{2}{5}$$
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