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: 78

Answer

$\ln\sqrt{x}$ grows slower than $\ln^2{x}$.

Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{\ln\sqrt{x}}{\ln^2 x}.$ 1) If it is equal to zero then $\ln\sqrt{x}$ grows slower than $\ln^2x$; 2) If it is equal to $\infty$ then $\ln\sqrt{x}$ grows faster than $\ln^2 x$; 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". Also, note that $\ln\sqrt{x}=\ln x^{1/2}=\frac{1}{2}\ln x$ so we have $$\lim_{x\to\infty}\frac{\ln\sqrt{x}}{\ln^2 x}=\lim_{x\to\infty}\frac{\frac{1}{2}\ln x}{\ln^2 x}=\lim_{x\to\infty}\frac{1}{2\ln x}=\left[\frac{1}{2\ln\infty}\right]=\left[\frac{1}{\infty}\right]=0, $$ and thus $\ln\sqrt{x}$ grows slower than $\ln^2{x}$.
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