Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise - Page 464: 23


$\lim\limits_{t \to \infty}(2)^{-t^{2}}=0$

Work Step by Step

Evaluate limit for $\lim\limits_{t \to - \infty}(2)^{-t^{2}}$ Consider $-t^{2} =u$ $\lim\limits_{t \to \infty}(2)^{-t^{2}}=\lim\limits_{x \to - \infty}(2)^{u}$ Since $\lim\limits_{a \to \infty}a^{x}=0$: when $a>1$ Here $a=2>1$ This implies $\lim\limits_{t \to \infty}(2)^{-t^{2}}=\lim\limits_{x \to -\infty}2^{x}=0$ Hence, $\lim\limits_{t \to \infty}(2)^{-t^{2}}=0$
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