Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.3 Basic Limit Laws - Exercises - Page 59: 33


See the example below.

Work Step by Step

Take $ f(x)=\ln x $ and $ g(x)=\ln \frac{1}{x}$. Now, the limits $$\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\ln x, \quad \lim\limits_{x \to 0}g(x)= \lim\limits_{x \to 0}\ln \frac{1}{x} $$ do not exist. Compare to the limit of the sum: $$\lim\limits_{x \to 0}f(x)+ \lim\limits_{x \to 0}g(x)= \lim\limits_{x \to 0}\ln x+\lim\limits_{x \to 0}\ln \frac{1}{x} =\lim\limits_{x \to 0}(\ln x+\ln \frac{1}{x} )$$ Since $\ln x+\ln \frac{1}{x} =\ln x\frac{1}{x}=\ln 1=0$, then $$\lim\limits_{x \to 0}f(x)+ \lim\limits_{x \to 0}g(x)= \lim\limits_{x \to 0}(\ln x+\ln \frac{1}{x} )=\lim\limits_{x \to 0}(\ln x \frac{1}{x} )=0.$$ Thus, the limit of the sum exists, even if the individual limits do not.
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