Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 10

Answer

See the proof below.

Work Step by Step

We have $$|f (x) − 3| = |3 − 3| = 0$$ for all $ x $, so when we make $ x $ sufficiently close to $3$, the result is sufficiently close to $0$. Hence, $$\lim_{x\rightarrow 5}3=3.$$
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