Answer
$-\frac{3}{4}$.
Work Step by Step
Since we have
$$\lim_{n\to \infty}a_n=\lim_{n\to \infty}\frac{4+n-3n^2}{4n^2+1}=\lim_{n\to \infty}\frac{(4/n^2)+(1/n)-3}{4+(1/n^2)}=-\frac{3}{4}$$
then the sequence $a_n$ converges to $-\frac{3}{4}$.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 537: 18
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