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: 25

Answer

$\lim _{x\rightarrow \infty }\dfrac {4x}{20x+1}=\dfrac {1}{5};\lim _{x\rightarrow -\infty }\dfrac {4x}{20x+1}=\dfrac {1}{5};y=\dfrac {1}{5}$

Work Step by Step

$\dfrac {4x}{20x+1}=\dfrac {4x+\dfrac {1}{5}-\dfrac {1}{5}}{20x+1}=\dfrac {1}{5}-\dfrac {1}{5\left( 20x+1\right) }=\dfrac {1}{5}-\dfrac {1}{100x+5}$ $$\Rightarrow \lim _{x\rightarrow \infty }\dfrac {4x}{20x+1}=\lim _{x\rightarrow \infty }\left( \dfrac {1}{5}-\dfrac {1}{100x+5}\right) =\dfrac {1}{5}-0=\dfrac {1}{5}$$ $$\Rightarrow \lim _{x\rightarrow -\infty }\dfrac {4x}{20x+1}=\lim _{x\rightarrow -\infty }\left( \dfrac {1}{5}-\dfrac {1}{100x+5}\right) =\dfrac {1}{5}+0=\dfrac {1}{5}$$ So horizontal asymptote will be $y=\frac{1}{5}$
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