University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Practice Exercises - Page 111: 5



Work Step by Step

$$\lim_{x\to0}\Big(\frac{4-g(x)}{x}\Big)=1$$ Here we cannot apply Quotient Rule because in the denominator, $\lim_{x\to0}x=0$. Instead, we would start with $g(x)$: $$g(x)=-(-g(x))=-(4-g(x)-4)=-\Big(\frac{4-g(x)}{x}\times x-4\Big)$$ Therefore, $$\lim_{x\to0}g(x)=\lim_{x\to0}\Big[-\Big(\frac{4-g(x)}{x}\times x-4\Big)\Big]$$ $$\lim_{x\to0}g(x)=-\Big[\lim_{x\to0}\Big(\frac{4-g(x)}{x}\Big)\times\lim_{x\to0}(x)-\lim_{x\to0}4\Big]$$ $$\lim_{x\to0}g(x)=-(1\times0-4)=-(-4)=4$$
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