Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2 - Page 70: 48

Answer

Using theorem, we know that limit does not exist for the quotient of two functions if the function in the denominator approaches to $0$. But according to the question $\lim_\limits{x\to a }\dfrac{f(x)}{g(x)}$ exists. So, there must be a common factor in both the functions which approaches to $0$ as $x\to a$. Since, this common factor from the numerator cancels the factor in the denominator, we get a finite value for the limit. Since, this factor is in $f(x)$ and this factor approaches to $0$ as $x\to a$. If $x\to a$, $f(x)$ also approaches to $0$. Hence, $\lim_\limits{x\to a}f(x)=0$

Work Step by Step

Using theorem, we know that limit does not exist for the quotient of two functions if the function in the denominator approaches to $0$. But according to the question $\lim_\limits{x\to a }\dfrac{f(x)}{g(x)}$ exists. So, there must be a common factor in both the functions which approaches to $0$ as $x\to a$. Since, this common factor from the numerator cancels the factor in the denominator, we get a finite value for the limit. Since, this factor is in $f(x)$ and this factor approaches to $0$ as $x\to a$. If $x\to a$, $f(x)$ also approaches to $0$. Hence, $\lim_\limits{x\to a}f(x)=0$
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