Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 77

Answer

$x^x$ grows faster than $(x/2)^x$.

Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{x^x}{(x/2)^x}.$ 1) If it is equal to zero then $x^x$ grows slower than $(x/2)^x$; 2) If it is equal to $\infty$ then $x^x$ grows faster than $(x/2)^x$; 3) If it is equal to some non zero constant then their growth rates are comparable. "LR" will stand for "Apply L'Hopital's rule": $$\lim_{x\to\infty}\frac{x^x}{\left(\frac{x}{2}\right)^x}=\lim_{x\to\infty}\left(\frac{x}{\frac{x}{2}}\right)^x=\lim_{x\to\infty}\left(2\right)^x=\left[2^\infty\right]=\infty,$$ and thus $x^x$ grows faster than $(x/2)^x$.
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