Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 72: 15


$$\lim _{t \rightarrow 0} \frac{4^{2 t}-1}{4^{t}-1}=2$$

Work Step by Step

Given $$\lim _{t \rightarrow 0} \frac{4^{2 t}-1}{4^{t}-1}$$ let $$ f(t) = \frac{4^{2 t}-1}{4^{t}-1}$$ Since, we have $$ f(0)=\frac{1-1}{1-1}=\frac{0}{0}$$ So, transform algebraically and cancel \begin{aligned} L&=\lim _{t \rightarrow 0} \frac{4^{2 t}-1}{4^{t}-1}\\ &=\lim _{t \rightarrow 0} \frac{(4^{ t})^2-1}{4^{t}-1}\\ &=\lim _{t \rightarrow 0} \frac{(4^{ t}-1)(4^t+1)}{4^{t}-1}\\ &= \lim _{h \rightarrow 0}(4^t+1)\\ &=1+1\\ &= 2 \end{aligned}
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