## Calculus 10th Edition

Published by Brooks Cole

# 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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