Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 1 - Limits and Continuity - 1.3 Limits At Infinity; End Behavior Of A Function - Exercises Set 1.3 - Page 79: 14

Answer

$$\frac{5}{2}$$

Work Step by Step

Given $$\lim _{x \rightarrow+\infty}\frac{5x^2-4x}{2x^2+3}$$ Then \begin{align*} \lim _{x \rightarrow+\infty}\frac{5x^2-4x}{2x^2+3}&=\lim _{x \rightarrow+\infty}\frac{\frac{5x^2}{x^2}-\frac{4x}{x^2}}{\frac{2x^2}{x^2}+\frac{3}{x^2}}\\ &=\lim _{x \rightarrow+\infty}\frac{5 -\frac{4x}{x^2}}{2 +\frac{3}{x^2}}\\ &=\frac{5}{2} \end{align*}
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