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: 40


We can define $h(2)=7$ that extends $h(t)$ to be continuous at $t=2$.

Work Step by Step

$$h(t)=\frac{t^2+3t-10}{t-2}$$ Currently, $h(t)$ is not defined at $t=2$, so $h(t)$ is not continuous at $t=2$ as well. For $h(t)$ to be continuous at $t=2$, we need to extend $h(t)$ to include a value of $h(2)$ so that $h(2)=\lim_{t\to2}h(t)$ So we need to calculate $\lim_{t\to2}h(t)$ first: $$\lim_{t\to2}h(t)=\lim_{t\to2}\frac{t^2+3t-10}{t-2}=\lim_{t\to2}\frac{(t-2)(t+5)}{t-2}=\lim_{t\to2}(t+5)$$ $$\lim_{t\to2}h(t)=2+5=7$$ So we can define $h(2)=7$ that extends $h(t)$ to be continuous at $t=2$.
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