Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 95

Answer

The solution is $$\lim_{n\to\infty}\frac{1+2+\cdots+n}{n^2}=\frac{1}{2}.$$

Work Step by Step

To solve this limit we will use the hint given in the problem $1+2+3+\cdots+n=\frac{n(n+1)}{2}$: $$\lim_{n\to\infty}\frac{1+2+\cdots+n}{n^2}=\lim_{n\to\infty}\frac{\frac{n(n+1)}{2}}{n^2}=\lim_{n\to\infty}\frac{n^2\left(1+\frac{1}{n}\right)}{2n^2}=\lim_{n\to\infty}\frac{1}{2}\left(1+\frac{1}{n}\right)=\left[\frac{1}{2}\left(1+\frac{1}{\infty}\right)\right]=\left[\frac{1}{2}\left(1+0\right)\right]=\frac{1}{2}.$$
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