University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 15

Answer

$3.99 \lt x \lt4.01$ $\delta = 0.01$

Work Step by Step

We solve the inequality: $|x+1 - 5|\lt 0.01$ $-0.01 \lt|x-4|\lt0.01$ $3.99 \lt|x| \lt4.01$ $3.99 \lt x \lt4.01$ Finding $\delta$: $|x-4|\lt \delta$ $-\delta+4\lt x \lt \delta+4$ $\delta=0.01$

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