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: 54

Answer

(a) Below, we can see a graph below of $f(x) = \lfloor cos~x \rfloor$ on the interval $-\pi \leq x \leq \pi$ (b) (i) $\lim\limits_{x \to 0}f(x) = 0$ (ii) $\lim\limits_{x \to (\pi/2)^-}f(x) = 0$ (iii) $\lim\limits_{x \to (\pi/2)^+}f(x) = -1$ (iv) $\lim\limits_{x \to \pi/2}f(x)$ does not exist (c) $\lim\limits_{x \to a}f(x)$ exists for all values of $a$ in the interval except $-\frac{\pi}{2}$ and $\frac{\pi}{2}$
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Work Step by Step

(a) Below, we can see a graph of $f(x) = \lfloor cos~x \rfloor$ on the interval $-\pi \leq x \leq \pi$ (b) We can evaluate each limit: (i) $\lim\limits_{x \to 0}f(x) = 0$ (ii) $\lim\limits_{x \to (\pi/2)^-}f(x) = 0$ (iii) $\lim\limits_{x \to (\pi/2)^+}f(x) = -1$ (iv) $\lim\limits_{x \to \pi/2}f(x)$ does not exist because $\lim\limits_{x \to (\pi/2)^-} f(x) \neq \lim\limits_{x \to (\pi/2)^+} f(x)$ (c) $\lim\limits_{x \to a}f(x)$ exists for all values of $a$ in the interval $-\pi \leq x \leq \pi$ except where the left limit is not equal to the right limit. Therefore, the limit exists for all values of $a$ except $-\frac{\pi}{2}$ and $\frac{\pi}{2}$
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