Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.5 - Continuity - Exercises 2.5 - Page 85: 40

Answer

Set $h( 2 )=7$

Work Step by Step

We see that $h(2) $ is not defined, and h is continuous everywhere else. We check if $\displaystyle \lim_{t\rightarrow 2}h(t)=L$ exists, and if it does, we can define the continuous extension of $h$ at $t=2$ as $H(t)=\left\{\begin{array}{ll} h(t) , & \text{if }t\neq 2\\ L, & \text{if }t=2 \end{array}\right.$ $L= \displaystyle \lim_{t\rightarrow 2}g(x)= \displaystyle \lim_{x\rightarrow 3}\frac{t^{2}+3t-10}{t-2}\quad $...factor the trinomial, $= \displaystyle \lim_{t\rightarrow 2}\frac{(t+5)(t-2)}{(t-2)}\quad $...cancel common term $= \displaystyle \lim_{t\rightarrow 2}(\mathrm{t}+5)$ ... evaluate directly, $L=7$ The limit exits at $t=2,$ so, if we redefine $h$, $h(t)=\left\{\begin{array}{ll} \dfrac{t^{2}+3t-10}{t-2} , & \text{if }t\neq 2\\ 7, & \text{if }t=2 \end{array}\right.$ it becomes continuous.
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