Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 76

Answer

$x^{10}\ln^{10}x$ grows slower than $x^{11}$.

Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{x^{10}\ln^{10}x}{x^{11}}.$ 1) If it is equal to zero then $x^{10}\ln^{10}x$ grows slower than $x^{11}$; 2) If it is equal to $\infty$ then $x^{10}\ln^{10}x$ grows faster than $x^{11}$; 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": $$L=\lim_{x\to\infty}\frac{x^{10}\ln^{10}x}{x^{11}}=\lim_{x\to\infty}\frac{\ln^{10}x}{x}=\lim_{x\to\infty}\left(\frac{\ln x}{x^{\frac{1}{10}}}\right)^{10}=\left(\lim_{x\to\infty}\frac{\ln x}{x^{\frac{1}{10}}}\right)^{10}=l^{10},$$ where we have denoted $l=\lim_{x\to\infty}\frac{\ln x}{x^{\frac{1}{10}}}$. Now we have $$l=\lim_{x\to\infty}\frac{\ln x}{x^{\frac{1}{10}}}=\left[\frac{\ln\infty}{\infty^{1/10}}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln x)'}{(x^{\frac{1}{10}})'}=\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{10}x^{-9/10}}=\lim_{x\to\infty}\frac{10}{x^{\frac{1}{10}}}=\left[\frac{10}{\infty^{1/10}}\right]=\left[\frac{10}{\infty}\right]=0.$$ Returning this into the initial limit we get $$L=l^{10}=0^{10}=0,$$ and thus $x^{10}\ln^{10}x$ grows slower than $x^{11}$.

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