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: 66

Answer

The function is continuous when $a=4.$

Work Step by Step

$\lim\limits_{x\to a^+}\dfrac{x^2-a^2}{x-a}=\lim\limits_{x\to a^+}\dfrac{(x+a)(x-a)}{(x-a)}$ $=\lim\limits_{x\to a^+}(x+a)=a^++a=2a.$ $\lim\limits_{x\to a^-}\dfrac{x^2-a^2}{x-a}=\lim\limits_{x\to a^-}\dfrac{(x-a)(x+a)}{(x-a)}$ $=\lim\limits_{x\to a^-}(x+a)=a^-+a=2a$ Since $\lim\limits_{x\to a^+}\dfrac{x^2-a^2}{x-a}=\lim\limits_{x\to a^-}\dfrac{x^2-a^2}{x-a}\to\lim\limits_{x\to a}\dfrac{x^2-a^2}{x-a}=2a.$ It is also evident from the given that $g(a)=8.$ For a function to be continuous $\lim\limits_{x\to a}g(x)=g(a)\to$ $2a=8\to a=4.$
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