Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises: 28

Answer

$$\lim _{x\rightarrow \infty }\dfrac {4x^{2}-7}{8x^{2}+5x+2}=\dfrac {1}{2};\lim _{x\rightarrow -\infty }\dfrac {4x^{2}-7}{8x^{2}+5x+2}=\dfrac {1}{2};y=\dfrac {1}{2}$$

Work Step by Step

$$\lim _{x\rightarrow \infty }\dfrac {4x^{2}-7}{8x^{2}+5x+2}=\dfrac {4-\dfrac {7}{x^{2}}}{8+\dfrac {5}{x}+\dfrac {2}{x^{2}}}=\dfrac {4-0}{8+0+0}=\dfrac {1}{2}$$ $$\lim _{x\rightarrow -\infty }\dfrac {4x^{2}-7}{8x^{2}+5x+2}=\dfrac {4-\dfrac {7}{x^{2}}}{8+\dfrac {5}{x}+\dfrac {2}{x^{2}}}=\dfrac {4-0}{8-0+0}=\dfrac {1}{2}$$ So the horizontal asymptote will be $y=\frac{1}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.