University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 278: 81



Work Step by Step

Consider $f(t)=\lim\limits_{t \to 0^{+}}\dfrac{e^t}{t}-\dfrac{1}{t}=\lim\limits_{t \to 0^{+}} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$ Simplify further: $\lim\limits_{t \to 0^{+}}\dfrac{e^t}{t}-\dfrac{1}{t}=\lim\limits_{t \to 0^{+}}\dfrac{e^t-1}{t}$ Thus, $f(0)=\dfrac{0}{0}$ This shows an Inderminate form of the limit, so apply L'Hospital's rule: $\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$ Thus, $\lim\limits_{t \to 0^{+}}\dfrac{e^t}{1}=1$
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