Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.4 Exercises: 64

Answer

The function is continuous when $a=4.$

Work Step by Step

$\lim\limits_{x\to0^-}g(x)=4[\lim\limits_{x\to0^-}\dfrac{\sin{x}}{x}]=4(1)=4.$ $\lim\limits_{x\to0^+}g(x)=a-2(0^+)=a.$ For the function to be continuous $\lim\limits_{x\to0^-}g(x)=\lim\limits_{x\to0^+}g(x)\to$ $a=4.$ Note: while $\lim\limits_{x\to0^-}\dfrac{\sin{x}}{x}$ was not studied, we know that $\lim\limits_{x\to0}\dfrac{\sin{x}}{x}=1$ and since this limit exists, this guarantees the existence of both one-sided limits and it also guarantees them being equal to $1.$
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