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: 72

Answer

$\ln x$ grows faster than $\ln(\ln x)$.

Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{\ln x}{\ln(\ln x)}.$ 1) If it is equal to zero then $\ln x$ grows slower than $\ln(\ln x)$; 2) If it is equal to $\infty$ then $\ln x$ grows faster than $\ln(\ln x)$; 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" $$\lim_{x\to\infty}\frac{\ln x}{\ln (\ln x)}=\left[\frac{\ln\infty}{\ln(\ln\infty)}\right]=\left[\frac{\infty}{\ln\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln x)'}{(\ln (\ln x))'}=\lim_{x\to\infty}\frac{(\ln x)'}{\frac{1}{\ln x}(\ln x)'}=\lim_{x\to\infty}\frac{1}{\frac{1}{\ln x}}=\lim_{x\to\infty}\ln x=\left[\ln\infty\right]=\infty,$$ and thus $\ln x$ grows faster than $\ln(\ln x)$.
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