University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 249: 88



Work Step by Step

L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\lim\limits_{h \to 0} f'(0)=\lim\limits_{h \to 0 }\dfrac{f(0+h)-f(0)}{h})$ and $\lim\limits_{h \to 0 }\dfrac{1/h}{e^{1/h^2}}=\dfrac{\infty}{\infty}$ This shows an indeterminate form of the limit, so we need to use L'Hospital's rule. $\lim\limits_{h \to 0} \dfrac{-1/h^2}{e^{1/h^2}(-2/h^2)}=\lim\limits_{h \to 0} \dfrac{h}{2} e^{-1/h^2}=0$
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