Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - True-False Quiz: 3


The statement is true.

Work Step by Step

$$\lim\limits_{x \to 1}\frac{x-3}{x^2+2x-4}=\frac{\lim\limits_{x \to 1}(x-3)}{\lim\limits_{x \to 1}(x^2+2x-4)}$$ Consider the denominator $$\lim\limits_{x \to 1}(x^2+2x-4)=1^2+2\times1-4=-1$$ Consdier the numerator $$\lim\limits_{x \to 1}(x-3)=1-3=-2$$ Therefore, both the limits exist. The limit of the denominator does not equal 0. So, it satisfies all the requirements of the quotient law, which states that $$\lim\limits_{x \to a}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x \to a}f(x)}{\lim\limits_{x \to a}g(x)}$$ In conclusion, the statement is true.
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