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: 15


See the proof below.

Work Step by Step

We have $$ |4x^2+2x+5-5| = |2x(x+1)| = 2 |2x+1| |x|=2 |2x+1| |x-0|$$ Since $|4x^2+2x+5 -5| $ is a multiple of $|x-0|$, then $|4x^2+2x+5 -5| $ is arbitrarily small whenever $ x $ is sufficiently close to $0$. Hence, we get $$\lim_{x\rightarrow 0}4x^2+2x+5=5.$$
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