## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Review - True-False Quiz: 2

#### Answer

The statement is false.

#### Work Step by Step

$$\lim\limits_{x \to 1}\frac{x^2+6x-7}{x^2+5x-6}=\frac{\lim\limits_{x \to 1}(x^2+6x-7)}{\lim\limits_{x \to 1}(x^2+5x-6)}$$ Consider the denominator $$\lim\limits_{x \to 1}(x^2+5x-6)=1^2+5\times1-6=0$$ The quotient law states that $$\lim\limits_{x \to a}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x \to a}f(x)}{\lim\limits_{x \to a}g(x)}$$ if $\lim\limits_{x \to a}g(x)\ne0$ (the limit of the denominator must not be 0) However, in the above statement, we see that the limit of the denominator equals 0. So, the statement is false.

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