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: 48

Answer

a. (i) 1 (ii) -1 (iii) DNE (iv) -1 (v) 1 (vi) DNE b. $lim_{x\to a}sgn\left(sinx\right)=DNE$ if $a=n\pi$ c. See graph
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Work Step by Step

a. $0^{+}$: means from right to 0 (small positive number) $0^{-}$: means from left to 0 (negative positive number) (i) $\lim _{x\to 0^+}sgn\left(sinx\right)=sgn\left(\lim \:_{x\to \:0^+}sinx\right)=sgn\left(0^+\right)=1$ (ii) $\lim _{x\to 0^-}sgn\left(sinx\right)=sgn\left(\lim \:_{x\to \:0^-}sinx\right)=sgn\left(0^-\right)=-1$ (iii) we have to look at (i) and (ii) $\lim _{x\to 0^+}sgn\left(sinx\right)\ne\lim _{x\to 0^-}sgn\left(sinx\right)$ $\lim _{x\to 0}sgn\left(sinx\right)=DNE$ (iv) $\lim _{x\to \:\pi \:^+\:}sgn\left(sinx\right)=sgn\left(\lim \:_{x\to\pi \:^+\:}sinx\right)=sgn\left(0^-\right)=-1$ (v) $\lim _{x\to \:\pi \:^-}sgn\left(sinx\right)=sgn\left(\lim \:_{x\to \pi \:^-}sinx\right)=sgn\left(0^+\right)=1$ (vi) we have to look (iv) and (v) $lim_{x\to \:\pi ^+\:}sgn\left(sinx\right)\ne lim_{x\to\pi \:^-}\:sgn\left(sinx\right)$ $lim_{x\to \:\pi}sgn\left(sinx\right)=DNE$ b. $\lim _{x\to a}sgn\left(sin(x)\right)$ does not exist when $a=0$ then $\lim _{x\to a}sgn\left(sin(x)\right)$ not exist when $sina=0$ $sin a=0$ when $a=n\pi$ $lim_{x\to a}sgn\left(sinx\right)=DNE$ if $a=n\pi$ c.See graph
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