Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.3 - The Precise Definition of a Limit - Exercises 2.3 - Page 66: 46


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Work Step by Step

Step 1. We can identify the quantities as: $f(x)=\frac{x^2-1}{x-1},c=1,L=2$. Since when $x\to1, x\ne1$, we can simplify the function as $f(x)=\frac{x^2-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1$ Step 2. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ when $0\lt |x-1|\lt\delta$, we have $|f(x)-2|\lt\epsilon$ or $|x-1|\lt\epsilon$ Step 3. The last inequality gives $-\epsilon\lt x-1 \lt\epsilon$ or $1-\epsilon\lt x\lt1+\epsilon$ Step 4. From the requirement for the value $\delta$ in step 2, we have $-\delta\lt x-1\lt\delta$ or $1-\delta\lt x\lt1+\delta$ Step 5. Compare the results from steps 4 and 3, we can set $\delta=\epsilon$ Step6. With the setting of the $\delta$ value, we can go backwards through the above steps to reach a conclusion that for all $x$ when $0\lt |x-1|\lt\delta$, we have $|f(x)-2|\lt\epsilon$ which proves the limit statement.
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