Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - Chapter Review Exercises - Page 94: 25



Work Step by Step

By substituting, we get the intermediate form $\frac{0}{0}$. So, by factorizing the numerator, we have $$ \lim _{x \rightarrow 0} \frac{4^{3x}-4^x}{4^x-1}=\lim _{x \rightarrow 0} \frac{4^x(4^x+1)(4^x-1)}{4^x-1}\\ =\lim _{x \rightarrow 0} 4^x(4^x+1)=4^0(4^0+1)=2. $$
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