Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.9 The Formal Definition of a Limit - Exercises - Page 93: 23


$$\lim_{x\to 2}x^{-2}=\frac{1}{4}$$

Work Step by Step

Let $\epsilon\gt 0$ be given. First, we bound $x^{-2}-\frac{1}{4}$ : $$ \left|x^{-2}-\frac{1}{4}\right|=\left|\frac{4-x^{2}}{4 x^{2}}\right|=|2-x|\left|\frac{2+x}{4 x^{2}}\right| $$ Let $\delta=\min \left(1, \frac{4}{5} \epsilon\right),$ and suppose $|x-2|\lt \delta .$ Since $\delta\lt 1,|x-2|\lt 1,$ so $1\lt x\lt 3 .$ This means that $4 x^{2}\gt 4$ and $|2+x|\lt 5,$ so that $\frac{2+x}{4 x^{2}}\lt \frac{5}{4} .$ We get: $$ \left|x^{-2}-\frac{1}{4}\right|=|2-x|\left|\frac{2+x}{4 x^{2}}\right|\lt \frac{5}{4}|x-2|\lt \frac{5}{4} \cdot \frac{4}{5} \epsilon=\epsilon $$ Hence $$\lim_{x\to 2}x^{-2}=\frac{1}{4}$$
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