Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Section 13.2 - Finding Limits Algebraically - 13.2 Exercises - Page 913: 25

Answer

$\lim_{h\to0}\dfrac{(2+h)^{2}-4}{h}=4$

Work Step by Step

$\lim_{h\to0}\dfrac{(2+h)^{2}-4}{h}$ Try to evaluate the limit using direct substitution: $\lim_{h\to0}\dfrac{(2+h)^{2}-4}{h}=\dfrac{(2+0)^{2}-4}{0}=\dfrac{0}{0}$ Indeterminate form The limit could not be evaluated using direct substitution. Factor the numerator of the function and simplify: $\lim_{h\to0}\dfrac{(2+h)^{2}-4}{h}=\lim_{h\to0}\dfrac{[(2+h)-2][(2+h)+2]}{h}=...$ $...=\lim_{h\to0}\dfrac{(h)(h+4)}{h}=\lim_{h\to0}(h+4)=...$ Try to evaluate the limit using direct substitution again: $\lim_{h\to0}(h+4)=0+4=4$
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