Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 410: 88



Work Step by Step

$\lim\limits_{h \to 0} f'(0)=\lim\limits_{h \to 0 }\dfrac{f(0+h)-f(0)}{h})$ and $\lim\limits_{h \to 0 }\dfrac{1/h}{e^{1/h^2}}=\dfrac{\infty}{\infty}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $\lim\limits_{h \to 0} \dfrac{\frac{-1}{h^2}}{e^{\frac{1}{h^2}}(\dfrac{-2}{h^2})}=\lim\limits_{h \to 0} (\dfrac{h}{2})( e^{-1/h^2})=0$ and $(0) \cdot (\infty)=0$
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