Find L=lim_(nrightarrow infty) (2n^2 +6n-1)/(4n^2 +2n+1). then determine and find.

Kyle Liu 2022-07-26 Answered
Find $L=\underset{n\to \mathrm{\infty }}{lim}\frac{2{n}^{2}+6n-1}{4{n}^{2}+2n+1}$. then determine and find .
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Answers (1)

Franklin Frey
Answered 2022-07-27 Author has 15 answers
find $L=\underset{n\to \mathrm{\infty }}{lim}﻿\frac{2{n}^{2}+6n-1}{4{n}^{2}+2n+1}=\frac{2}{4}=\frac{1}{2}⇒L=\frac{1}{2}$
then determine $\left\{{ϵ}_{n}\right\}=\left\{\frac{1}{2}-{x}_{n}\right\}$ and find
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2}-\underset{n\to \mathrm{\infty }}{lim}{x}_{n}=\frac{1}{2}-\underset{n\to \mathrm{\infty }}{lim}{x}_{n}$
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