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: 70

Answer

$x^2\ln x$ grows faster than $\ln^2x$.

Work Step by Step

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