Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises: 10

Answer

(a) The equation has one side that is not defined with $x=2$, while the other side is still defined with $x=2$. (b) The equation is correct because when computing the limit as $x$ approaches $2$, we never actually consider the situation when $x=2$.

Work Step by Step

(a) Consider the equation. $\frac{x^2+x-6}{x-2}=x+3$ We see that - In the left side of the equation, $x\ne2$ or else, the left side would not be defined. - In the right side of the equation, $x+3$ is defined with $x=2$. Therefore, the equation has one side that is not defined with $x=2$, while the other side is still defined with $x=2$. (b) We see that $\lim\limits_{x\to 2}\frac{x^2+x-6}{x-2}$ $=\lim\limits_{x\to 2}\frac{(x-2)(x+3)}{x-2}$ $=\lim\limits_{x\to 2} (x+3)$ While the left-hand side is undefined for $x=2$, when we compute the limit as $x$ approaches $2$, we never consider the situation when $x=2$, but only the value of $x$ near $2$. Also, both sides are defined any $x\ne2$. Therefore, the equation is still correct.
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