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: 79

Answer

$e^{x^2}$ grows faster than $e^{10x}$.

Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{e^{x^2}}{e^{10x}}.$ 1) If it is equal to zero then $e^{x^2}$ grows slower than $e^{10x}$; 2) If it is equal to $\infty$ then $e^{x^2}$ grows faster than $e^{10x}$; 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": $$\lim_{x\to\infty}\frac{e^{x^2}}{e^{10x}}=\lim_{x\to\infty}e^{x^2-10x}=e^{\lim_{x\to\infty}(x^2-10x)}=[e^\infty]=\infty,$$ and thus $e^{x^2}$ grows faster than $e^{10x}$. NOTE: We used that $\lim_{x\to\infty}(x^2-10x)=\lim_{x\to\infty}x^2(1-10/x)=[\infty^2(1-0)]=\infty$.
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