Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 104: 52

Answer

(a). Evaluate each limit: (i) $\lim\limits_{x \to 1^{-}}$ $g(x) = 1$ (ii) $\lim\limits_{x \to 1}$ $g(x) = 1$ (iii) $g(1) = 3$ (iv) $\lim\limits_{x \to 2^{-}}$ $g(x) = -2$ (v) $\lim\limits_{x \to 2^{+}}$ $g(x) = -1$ (vi) $\lim\limits_{x \to 2}$ $g(x) =$ Does not exist b. Graph
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Work Step by Step

(i) $\lim\limits_{x \to 1^{-}}$ $g(x)$ $\lim\limits_{x \to 1^{-}}$ $x = 1$ (ii) $\lim\limits_{x \to 1}$ $g(x)$ $\lim\limits_{x \to 1} (2-x^{2})$ $ \lim\limits_{x \to 1} (2 - 1^{2}) = 1$ (iii) $g(1)$ $g(1) = 3$ (iv) $\lim\limits_{x \to 2^{-}} g(x)$ $\lim\limits_{x \to 2^{-}} (2-x^{2})$ $\lim\limits_{x \to 2^{-}} (2-2^{2}) = (2 - 4) = -2$ (v) $\lim\limits_{x \to 2^{+}} g(x)$ $\lim\limits_{x \to 2^{+}} (x-3)$ $\lim\limits_{x \to 2^{+}} (2 - 3) = -1$ (vi) $\lim\limits_{x \to 2} g(x)$ The limit does not exists because $\lim\limits_{x \to 2^{+}} g(x)$ $\ne$ $\lim\limits_{x \to 2^{-}} g(x)$
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