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: 73

Answer

$100^x$ grows slower than $x^x$.

Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{100^x}{x^x}.$ 1) If it is equal to zero then $100^x$ grows slower than $x^x$; 2) If it is equal to $\infty$ then $100^x$ grows faster than $x^x$; 3) If it is equal to some non zero constant then their growth rates are comparable. $$\lim_{x\to\infty}\frac{100^x}{x^x}=\lim_{x\to\infty}\left(\frac{100}{x}\right)^x=\left[\left(\frac{100}{\infty}\right)^\infty\right]=\left[0^\infty\right]=0,$$ and thus $100^x$ grows slower than $x^x$.
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