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: 41


We can define $f(1)=\frac{3}{2}$ that extends $f(s)$ to be continuous at $s=1$.

Work Step by Step

$$f(s)=\frac{s^3-1}{s^2-1}$$ Currently, $f(s)$ is not defined at $s=1$, so $f(s)$ is not continuous at $s=1$ as well. For $f(s)$ to be continuous at $s=1$, we need to extend $f(s)$ to include a value of $f(1)$ so that $f(1)=\lim_{s\to1}f(s)$ So we need to calculate $\lim_{s\to1}f(s)$ first: $$\lim_{s\to1}f(s)=\lim_{s\to1}\frac{s^3-1}{s^2-1}=\lim_{s\to1}\frac{(s-1)(s^2+s+1)}{(s-1)(s+1)}=\lim_{s\to1}\frac{s^2+s+1}{s+1}$$ $$\lim_{s\to1}f(s)=\frac{1^2+1+1}{1+1}=\frac{3}{2}$$ So we can define $f(1)=\frac{3}{2}$ that extends $f(s)$ to be continuous at $s=1$.
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