Calculus (3rd Edition)

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

Chapter 2 - Limits - Chapter Review Exercises - Page 95: 69


$$ \lim _{x \rightarrow 0} x\left[\frac{1}{x}\right]=1 $$

Work Step by Step

Let $y$ be any real number. From the definition of the greatest integer function, it follows that $y-1\lt [y] \leq y$, with equality holding if and only if $y$ is an integer. If $x \neq 0,$ then $\frac{1}{x}$ is a real number, so $$ \frac{1}{x}-1\lt \left[\frac{1}{x}\right] \leq \frac{1}{x} $$ Upon multiplying this inequality through by $x,$ we find $$ 1-x\lt x\left[\frac{1}{x}\right] \leq 1 $$ Since $$\lim _{x \rightarrow 0}(1-x)=\lim _{x \rightarrow 0} 1=1$$ Thus, it follows from the Squeeze Theorem that $$ \lim _{x \rightarrow 0} x\left[\frac{1}{x}\right]=1 $$
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