Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises: 31

Answer

The sequence converges to $\sqrt{4} = 2$.

Work Step by Step

We can take the limit inside of the square root. Since the highest degree of $n$ in the numerator is 2 and the highest degree in the denominator is 2, the limit as $n$ approaches infinity of the inner function will be that of $\frac{4n^2}{n^2} = 4$. Hence the sequence converges to $\sqrt{4} = 2$.
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