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: 69

Answer

$x^{10}$ grows slower than $e^{0.01x}$.

Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{x^{10}}{e^{0.01x}}.$ 1) If it is equal to zero then $x^{10}$ grows slower than $e^{0.01x}$; 2) If it is equal to $\infty$ then $x^{10}$ grows faster than $e^{0.01x}$; 3) If it is equal to some constant the their growth rates are comparable. $$L=\lim_{x\to\infty}\frac{x^{10}}{e^{0.01x}}=\lim_{x\to\infty}\left(\frac{x}{e^{\frac{0.01x}{10}}}\right)^{10}=\left(\lim_{x\to\infty}\frac{x}{e^{0.001x}}\right)^{10}=l^{10},$$ where we denoted $l=\lim_{x\to\infty}\frac{x}{e^{0.001x}}.$ Now we have $$\lim_{x\to\infty}\frac{x}{e^{0.001x}}=\left[\frac{\infty}{e^\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(x)'}{(e^{0.001x})'}=\lim_{x\to\infty}\frac{1}{0.001e^{0.001x}}=\left[\frac{1}{0.001e^\infty}\right]=\left[\frac{1}{\infty}\right]=0.$$ Now returning to the initial limit we have $$L=l^{10}=0^{10}=0,$$ and thus $x^{10}$ grows slower than $e^{0.01x}$.
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