Using the definition of a convergent sequence, prove:
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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*You cannot use any theorems about convergent sequences*
Using the definition of a convergent sequence, prove:
![The expression shown in the image is a mathematical limit:
\[
\lim_{{n \to \infty}} \frac{n}{2n-1} = \frac{1}{2}
\]
This limit describes the behavior of the fraction \(\frac{n}{2n-1}\) as \(n\) approaches infinity. As \(n\) becomes very large, the value of the fraction approaches \(\frac{1}{2}\). This is because the highest degree terms in the numerator and the denominator dominate, simplifying the expression to \(\frac{1}{2}\) in the limit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbaa9c753-6b68-41b4-8837-61aa8aa0a19b%2Fc1e01997-b6c5-408f-9a87-d26909ac2cfb%2Fp52jso_processed.png&w=3840&q=75)
Transcribed Image Text:The expression shown in the image is a mathematical limit:
\[
\lim_{{n \to \infty}} \frac{n}{2n-1} = \frac{1}{2}
\]
This limit describes the behavior of the fraction \(\frac{n}{2n-1}\) as \(n\) approaches infinity. As \(n\) becomes very large, the value of the fraction approaches \(\frac{1}{2}\). This is because the highest degree terms in the numerator and the denominator dominate, simplifying the expression to \(\frac{1}{2}\) in the limit.
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