Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 44

Answer

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Work Step by Step

Given $$ \lim _{x \rightarrow \infty} \sqrt{x} e^{-x / 2}$$ Rewrite \begin{aligned} \lim _{x \rightarrow \infty} \sqrt{x} e^{-x / 2}&= \lim _{x\to \infty \:}\left(\frac{1}{e^{\frac{x}{2}}}\sqrt{x}\right)\\ &= \lim _{x\to \infty \:}\left(\sqrt{\frac{x}{e^x}}\right)\\ &= \sqrt{\lim _{x\to \infty \:}\left(\frac{x}{e^x}\right)} \end{aligned} Apply L'Hopital's rule for \begin{aligned} \lim _{x\to \infty \:}\left(\frac{x}{e^x}\right)&= \lim _{x\to \infty \:}\left(\frac{1}{e^x}\right)\\ &=\frac{1}{\infty}=0 \end{aligned} Then \begin{aligned} \lim _{x\to \infty \:}\left(\sqrt{\frac{x}{e^x}}\right)&= \sqrt{0}=0 \end{aligned}
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