Answer
$$12$$
Work Step by Step
\begin{align*}
\lim _{x \rightarrow 0} \frac{(2+x)^{3}-8}{x}&=\lim _{x \rightarrow 0} \frac{\left(x^{3}+6 x^{2}+12 x+8\right)-8}{x}\\
&=\lim _{x \rightarrow 0}\left(x^{2}+6 x+12\right)\\
&=12
\end{align*}
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 2: Limits and Continuity - Practice Exercises - Page 101: 16
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