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: 74

Answer

$x^2\ln x$ grows slower than $x^3$.

Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{x^2\ln x}{x^3}.$ 1) If it is equal to zero then $x^2\ln x$ grows slower than $x^3$; 2) If it is equal to $\infty$ then $x^2\ln x$ grows faster than $x^3$; 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": $$\lim_{x\to\infty}\frac{x^2\ln x}{x^3}=\lim_{x\to\infty}\frac{\ln x}{x}=\left[\frac{\ln\infty}{\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln x)'}{(x)'}=\lim_{x\to\infty}\frac{\frac{1}{x}}{1}=\lim_{x\to\infty}\frac{1}{x}=\left[\frac{1}{\infty}\right]=0,$$ and thus $x^2\ln x$ grows slower than $x^3$.
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