Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 82: 11



Work Step by Step

Divide the numerator and denominator by $x$, which is the highest power of $x$ occurring in the denominator. Then, we get $\lim\limits_{x \to \infty}\frac{7x-9}{4x+3}=\lim\limits_{x \to \infty}\frac{\frac{7x}{x}-\frac{9}{x}}{\frac{4x}{x}+\frac{3}{x}}=\lim\limits_{x \to \infty}\frac{7-\frac{9}{x}}{4+\frac{3}{x}}=\frac{7-0}{4+0}=\frac{7}{4}$
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