Answer
$$\lim_{x\to0}\frac{(2+x)^3-8}{x}=12$$
Work Step by Step
$$A=\lim_{x\to0}\frac{(2+x)^3-8}{x}=\lim_{x\to0}\frac{8+3\times2x^2+3\times4x+x^3-8}{x}$$
$$A=\lim_{x\to0}\frac{6x^2+12x+x^3}{x}=\lim_{x\to0}(x^2+6x+12)$$
$$A=0^2+6\times0+12=12$$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 2 - Practice Exercises - Page 111: 16
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