Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises - Page 125: 33

Answer

The function is discontinuous at $x = 0$.

Work Step by Step

$\lim\limits_{x \to 0^{-}}\frac{1}{1+e^{\frac{1}{x}}} = \frac{1}{1+e^{-\infty}} = \frac{1}{1+0} = 1$ $\lim\limits_{x \to 0^{+}}\frac{1}{1+e^{\frac{1}{x}}} = \frac{1}{1+e^{\infty}} = \frac{1}{1+\infty} = 0$ Since both limits are not the same, we can conclude that $y = \frac{1}{1+e^{\frac{1}{x}}}$ is not continuous at $x = 0$.
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