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: 31


$$\lim _{t \rightarrow 2} \frac{2^{2 t}+2^{t}-20}{2^{t}-4}=9$$

Work Step by Step

Given $$\lim _{t \rightarrow 2} \frac{2^{2 t}+2^{t}-20}{2^{t}-4}$$ let $$ f(t) = \frac{2^{2 t}+2^{t}-20}{2^{t}-4}$$ Since, we have $$ f(2)= \frac{16+4-20}{4-4}$=\frac{0}{0}$$ So, transform algebraically and cancel \begin{aligned} L&=\lim _{t \rightarrow 2} \frac{2^{2 t}+2^{t}-20}{2^{t}-4}\\ &= \lim _{t \rightarrow 2} \frac{(2^{ t})^2+2^{t}-20}{2^{t}-4}\\ &=\lim _{t \rightarrow 2} \frac{(2^{ t}-4)(2^t+5) }{2^{t}-4}\\ &=\lim _{t \rightarrow 2} (2^t+5) \\ &= \frac{10+1}{5+5}\\ &=2^2+5\\ &=9 \end{aligned}
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