Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Questions to Guide Your Review - Page 635: 4

Answer

See below.

Work Step by Step

Theorem 4. Reworded in plain language, it states that: if we find a FUNCTION $f(x)$ such that, after a certain index $n_{0},$ all terms of the sequence are in fact function values $f(n)$ of the term's index, then the existence of a limit $\displaystyle \lim_{x\rightarrow\infty}f(x)=L$ implies that $\displaystyle \lim_{n\rightarrow\infty}a_{n}=L.$ $\displaystyle \lim_{x\rightarrow\infty}f(x)$ is the limit of a function $\displaystyle \lim_{n\rightarrow\infty}a_{n}$ is the limit of a sequence, if they exist. L'Hopital's rule applies to limits of FUNCTIONS, and we can use it to find the limit of a SEQUENCE. Example: Define the general term of $\{a_{n}\}$ as $a_{n}=e^{-n}\ln(n)$ Then, the function $f(x)=e^{-x}\ln x$ fits the requirements of the theorem. We can rewrite $f(x)$ as $ f(x)=\displaystyle \frac{\ln x}{e^{x}}, \quad$ and searching for the limit $\displaystyle \lim_{x\rightarrow\infty}f(x)$ =$\displaystyle \lim_{x\rightarrow\infty}\frac{\ln x}{e^{x}}\quad$ we have $\displaystyle \frac{\infty}{\infty}$ so we can apply L'Hopital's rule (differentiate up and down): $=\displaystyle \lim_{x\rightarrow\infty}\frac{[\ln x]'}{[e^{x}]'}=\lim_{x\rightarrow\infty}\frac{1/x}{e^{x}}=\lim_{x\rightarrow\infty}\frac{1}{xe^{x}}=$ The numerator is constant, and the denominator is the product of two large numbers the ratio approaches 0 $=0$ Now, by Theorem 4, since $\displaystyle \lim_{x\rightarrow\infty}f(x)=0$, it follows that $\displaystyle \lim_{n\rightarrow\infty}a_{n}=0$
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