Answer
$$0$$
Work Step by Step
Since, $\{ b^n \}\lt\lt \{ n! \}\ $ for $ b \gt 1$
Therefore, we can write
$\lim\limits_{n \to \infty}\dfrac{8^n}{n!}=0$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 8 - Sequences and Infinite Series - Review Exercises - Page 658: 3
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