Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.2 Definitions of Limits - 2.2 Exercises - Page 68: 35

Answer

$\approx 1$

Work Step by Step

We have to evaluate the limit: $\lim\limits_{h \to 0} f(h)$, where $f(h)=\dfrac{\ln (1+h)}{h}$ We compute the value of the function $f$ for values of $h$ close to 0: $f(0.01)\approx 0.99503309$ $f(0.001)\approx 0.99950033$ $f(0.0001)\approx 0.99995$ $f(-0.01)\approx 1.0050336$ $f(-0.001)\approx 1.0005003$ $f(-0.0001)\approx 1.00005$ We got $\lim\limits_{h \to 0} f(h)\approx 1$
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