University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.5 - Continuity - Exercises - Page 95: 42


We can define $g(4)=\frac{8}{5}$ that extends $g(x)$ to be continuous at $x=4$.

Work Step by Step

$$g(x)=\frac{x^2-16}{x^2-3x-4}=\frac{x^2-16}{(x-4)(x+1)}$$ Currently, $g(x)$ is not defined at $x=4$, so $g(x)$ is not continuous at $x=4$ as well. For $g(x)$ to be continuous at $x=4$, we need to extend $g(x)$ to include a value of $g(4)$ so that $g(4)=\lim_{x\to4}g(x)$ So we need to calculate $\lim_{x\to4}g(x)$ first: $$\lim_{x\to4}g(x)=\lim_{x\to4}\frac{x^2-16}{(x-4)(x+1)}=\lim_{x\to4}\frac{(x-4)(x+4)}{(x-4)(x+1)}=\lim_{x\to4}\frac{x+4}{x+1}$$ $$\lim_{x\to4}g(x)=\frac{4+4}{4+1}=\frac{8}{5}$$ So we can define $g(4)=\frac{8}{5}$ that extends $g(x)$ to be continuous at $x=4$.
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