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: 53


(a) (i) $\lim\limits_{x \to -2^+}\lfloor{x}\rfloor = -2$ (ii) $\lim\limits_{x \to -2}\lfloor{x}\rfloor $ does not exist. (iii)$\lim\limits_{x \to -2.4}\lfloor{x}\rfloor = -3$ (b) (i) $\lim\limits_{x \to n^-}\lfloor{x}\rfloor = n-1$ (ii) $\lim\limits_{x \to n^+}\lfloor{x}\rfloor = n$ (c) $\lim\limits_{x \to a}\lfloor{x}\rfloor$ exists if $a$ is not an integer. If $a$ is an integer, then the limit does not exist.

Work Step by Step

(a) (i) $\lim\limits_{x \to -2^+}\lfloor{x}\rfloor = \lim\limits_{x \to -2^+} -2 = -2$ (ii) $\lim\limits_{x \to -2^-}\lfloor{x}\rfloor = \lim\limits_{x \to -2^-} -3 = -3$ Since $\lim\limits_{x \to -2^-}\lfloor{x}\rfloor \neq \lim\limits_{x \to -2^+}\lfloor{x}\rfloor$, then $\lim\limits_{x \to -2}\lfloor{x}\rfloor $ does not exist. (iii)$\lim\limits_{x \to -2.4}\lfloor{x}\rfloor = \lim\limits_{x \to -2.4} -3 = -3$ (b) Suppose that $n$ is an integer. (i) $\lim\limits_{x \to n^-}\lfloor{x}\rfloor = \lim\limits_{x \to n^-} (n-1) = n-1$ (ii) $\lim\limits_{x \to n^+}\lfloor{x}\rfloor = \lim\limits_{x \to n^+} n = n$ (c) If $a$ is an integer, then $\lim\limits_{x \to a^-}\lfloor{x}\rfloor \neq \lim\limits_{x \to a^+}\lfloor{x}\rfloor$. Therefore, $\lim\limits_{x \to a}\lfloor{x}\rfloor$ does not exist if $a$ is an integer. If $a$ is not an integer, then $\lim\limits_{x \to a^-}\lfloor{x}\rfloor = \lim\limits_{x \to a^+}\lfloor{x}\rfloor = \lfloor{a}\rfloor$ Therefore, $\lim\limits_{x \to a}\lfloor{x}\rfloor$ exists if $a$ is not an integer.
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