Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 312: 44


$$\lim_{x\to\infty}\sqrt xe^{-x/2}=0$$

Work Step by Step

$$A=\lim_{x\to\infty}\sqrt xe^{-x/2}$$ $$A=\lim_{x\to\infty}\frac{\sqrt x}{e^{x/2}}$$ $$A=\lim_{x\to\infty}\frac{\sqrt x}{\sqrt{e^x}}$$ $$A=\lim_{x\to\infty}\sqrt{\frac{x}{e^x}}$$ $$A=\sqrt{\lim_{x\to\infty}\frac{x}{e^x}}$$ As $x\to\infty$, $e^x\to\infty$. We have an indeterminate form of $\infty/\infty$, which can dealt with by L'Hospital's Rule: $$A=\sqrt{\lim_{x\to\infty}\frac{x'}{(e^x)'}}$$ $$A=\sqrt{\lim_{x\to\infty}\frac{1}{e^x}}$$ We know from above that $\lim_{x\to\infty}(e^x)=\infty$, therefore, $\lim_{x\to\infty}\frac{1}{e^x}=0$. So, $$A=\sqrt0=0$$
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